Overview
Limits of a function
Let \(f\) be a function defined in a domain which we take to be an interval, say, I. We shall study the concept of limit of \(f\) at a point ‘ \(a\) ‘ in \(I\).
We say \(\lim _{x \rightarrow a^{-}} f(x)\) is the expected value of \(f\) at \(x=a\) given the values of \(f\) near to the left of \(a\). This value is called the left hand limit of \(f\) at \(a\).
We say \(\lim _{x \rightarrow a^{+}} f(x)\) is the expected value of \(f\) at \(x=a\) given the values of \(f\) near to the right of \(a\). This value is called the right hand limit of \(f\) at \(a\).
If the right and left hand limits coincide, we call the common value as the limit of \(f\) at \(x=a\) and denote it by \(\lim _{x \rightarrow a} f(x)\).
Some properties of limits
Let \(f\) and \(g\) be two functions such that both \(\lim _{x \rightarrow a} f(x)\) and \(\lim _{x \rightarrow a} g(x)\) exist. Then
Limits of polynomials and rational functions
If \(f\) is a polynomial function, then \(\lim _{x \rightarrow a} f(x)\) exists and is given by
\(
\lim _{x \rightarrow a} f(x)=f(a)
\)
An Important limit
An important limit which is very useful and used in the sequel is given below:
\(
\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}
\)
Remark The above expression remains valid for any rational number provided ‘ \(a\) ‘ is positive.
Limits of trigonometric functions
To evaluate the limits of trigonometric functions, we shall make use of the following limits which are given below:
Derivatives
Suppose \(f\) is a real valued function, then
\(
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \dots(1)
\)
is called the derivative of \(f\) at \(x\), provided the limit on the R.H.S. of (1) exists.
Algebra of derivative of functions
Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules of derivatives to follow closely that of limits as given below:
Let \(f\) and \(g\) be two functions such that their derivatives are defined in a common domain. Then:
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\(
\lim _{x \rightarrow 0} \frac{\sin x}{x(1+\cos x)} \text { is equal to }
\)
(b) is the correct answer, we have
\(
\lim _{x \rightarrow 0} \frac{\sin x}{x(1+\cos x)}=\lim _{x \rightarrow 0} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{x 2 \cos ^2 \frac{x}{2}}
\)
\(=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\tan \frac{x}{2}}{\frac{x}{2}}=\frac{1}{2}\)
\(
\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x} \text { is equal to }
\)
(a) is the correct answer, since
\(
\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x}=\lim _{y \rightarrow 0} \frac{1-\sin \frac{\pi}{2}-y}{\cos \frac{\pi}{2}-y} \quad \text { taking } \frac{\pi}{2}-x=y
\)
\(
\begin{aligned}
&=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sin y}=\lim _{y \rightarrow 0} \frac{2 \sin ^2 \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}} \\
&=\lim _{y \rightarrow 0} \tan \frac{y}{2}=0
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{|x|}{x} \text { is equal to }\)
(d) is the correct answer, since
\(\text { R.H.S }=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\frac{x}{x}=1\)
and
\(\text { L.H.S }=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x}=\frac{-x}{x}=-1\)
\(\lim _{x \rightarrow 1}[x-1] \text {, where [.] is greatest integer function, is equal to }\)
(d) is the correct answer, since
\(
\text { R.H.S }=\lim _{x \rightarrow 1^{+}}[x-1]=0
\)
and
L.H.S \(=\lim _{x \rightarrow 1^{-}}[x-1]=-1\)
\(\lim _{x \rightarrow 0} x \sin \frac{1}{x} \text { is equals to }\)
(a) is the correct answer, since
\(\lim _{x \rightarrow 0} x=0[latex] and [latex]-1 \leq \sin \frac{1}{x} \leq 1\), by Sandwitch Theorem, we have
\(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0\)
\(\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}, n \in \mathbf{N} \text {, is equal to }\)
(c) is the correct answer. As \(\lim _{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}\)
\(
=\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2}=\lim _{x \rightarrow \infty} \frac{1}{2} 1+\frac{1}{n}=\frac{1}{2}
\)
\(
\text { If } f(x)=x \sin x \text {, then } f^{\prime}Â \frac{\pi}{2} \text { is equal to }
\)
(b) is the correct answer. As \(f^{\prime}(x)=x \cos x+\sin x\)
So,
\(
f^{\prime} \frac{\pi}{2}=\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2}=1
\)
\(
\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi} \text { is }
\)
\(
\lim _{x \rightarrow \pi} \frac{\sin (x)}{x-\pi}=\lim _{x \rightarrow \pi} \cos (x)=-1
\)
\(
\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}
\)
(a)Â
\(
\begin{aligned}
&\lim _{x \rightarrow 0} \frac{x^2 \operatorname{Cos} x}{1-\operatorname{Cos} x} \\
&=\lim _{x \rightarrow 0} \frac{\left(x^2 \operatorname{Cos} x\right)}{\left(2 \operatorname{Sin}^2 \frac{x}{2}\right)} \\
&\because 1-\operatorname{Cos} x=2 \operatorname{Sin}^2 \frac{x}{2} \\
&=\lim _{x \rightarrow 0} \frac{\frac{x^2}{4} \times 4 \operatorname{Cos} x}{2 \operatorname{Sin}^2 \frac{x}{2}} \\
&=\lim _{x \rightarrow 0=>\frac{x}{2} \rightarrow 0} \frac{\left(\frac{x}{2}\right)^2 \cdot 2 \operatorname{Cos} x}{\frac{\operatorname{Sin}^2 x}{2}} \\
&=\lim _{x \rightarrow 0}\left(\frac{\frac{x}{2}}{\operatorname{Sin} \frac{x}{2}}\right) \cdot 2 \operatorname{Cos} x \\
&=\lim _{x \rightarrow 0} 2 \operatorname{Cos} x=2 \times 1=2
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} \text { is }\)
(a) If we assume the question was: \(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}\)
We have an indeterminate limit of the form \(\frac{0}{0}\)
Hence, L’Hopital’s rule applies.
\(
\therefore \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left((1+x)^n-1\right)}{\frac{d}{d x}(x)}
\)
\(=\lim _{x \rightarrow 0} \frac{n(1+x)^{n-1} \times 1-0}{1}\) [Power rule and chain rule]
\(=\lim _{x \rightarrow 0} n(1+x)^{n-1}\)
\(
=n \times 1^{n-1}=n
\)
\(\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}\) is
(b) We have, \(\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}\)
\(
\begin{aligned}
&=\lim _{x \rightarrow 1}\left\{\frac{x^m-1^m}{x-1} \times \frac{x-1}{x^n-1^n}\right\} \\
&=\lim _{x \rightarrow 1}\left\{\frac{x^m-1^m}{x-1}\right\} \times \lim _{x \rightarrow 1}\left\{\frac{x-1}{x^n-1^n}\right\} \\
&=\frac{m(1)^{m-1}}{n(1)^{n-1}}=\frac{m}{n}
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} \frac{1-\cos 4 \mathrm{x}}{1-\cos 6 \mathrm{x}} \text { is }
\)
(a)
\(
\begin{aligned}
&\lim _{(x \rightarrow 0)}((1-\cos 4 \mathrm{x}) /(1-\cos 6 \mathrm{x})) \\
&=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^2 2 x}{2 \sin ^2 3 x}\right) \\
&=\lim _{x \rightarrow 0}\left(\frac{\frac{\sin ^2 2 x}{4 x^2} \times 4 x^2}{\frac{\sin ^2 3 x}{9 x^2} \times 9 x^2}\right) \\
&=\frac{4}{9}
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} \frac{\operatorname{cosec} x-\cot x}{x}
\)
(c)
\(
\begin{aligned}
&\operatorname{cosecx}-\cot x=\frac{(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)}{\operatorname{cosec} x+\cot x}=\frac{\operatorname{cosec}^2 x-\cot ^2 x}{\operatorname{cosec} x+\cot x} \\
&=\frac{1}{\operatorname{cosec} x+\cot x}=\frac{\sin x}{1+\cos x} \\
&\text { So, } \\
&l=\lim _{x \rightarrow 0} \frac{\cos e c x-\cot x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x(1+\cos x)} \\
&l=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{1+\cos x}=(1)\left(\frac{1}{1+\cos 0}\right)=\frac{1}{2}
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}} \text { is }
\)
(c) Given, \(\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}\) \(=\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}} \cdot \frac{\sqrt{x+1}+\sqrt{1-x}}{\sqrt{x+1}+\sqrt{1-x}}\) \(=\lim _{x \rightarrow 0} \frac{\sin (\sqrt{x+1}+\sqrt{1-x})}{x+1-1+x}\) \(=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x} \lim _{x \rightarrow 0}(\sqrt{x+1}+\sqrt{1-x})\) \(=\frac{1}{2.1} \cdot 2=1\left[\therefore \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\)
\(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1}\)
(d)
\(
\begin{aligned}
&\lim _{x \rightarrow \frac{\pi}{4}} \frac{1+\tan ^2 x-2}{\tan x-1} \text { where } \sec ^2 x=1+\tan ^2 x \\
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^2 x-1}{\tan x-1} \\
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{(\tan x-1)(\tan x+1)}{\tan x-1} \\
&=\lim _{x \rightarrow \frac{\pi}{4}}(\tan x+1) \\
&=\tan \frac{\pi}{4}+1 \\
&=1+1=2
\end{aligned}
\)
\(
\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3} \text { is }
\)
(b)
\(
\begin{aligned}
&\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\
&=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)} \\
&=\lim _{x \rightarrow 1} \frac{2 x-3}{(2 x+3)(\sqrt{x}+1)} \\
&=\frac{2-3}{(2+3)(1+1)} \\
&=-\frac{1}{10}
\end{aligned}
\)
\(
\text { If } f(x)=\left\{\begin{array}{cc}
\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\
0 & ,[x]=0
\end{array}\right.
\)
where [.] denotes the greatest integer function,Â
\(
\text { then } \lim _{x \rightarrow 0} f(x) \text { is equal to }
\)
(d)
\(
\begin{aligned}
&\text { Since, } f(x)=\left\{\begin{array}{cc}
\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\
0, & {[x]=0}
\end{array}\right. \\
&\Rightarrow f(x)=\left\{\begin{array}{cc}
\frac{\sin [x]}{[x]}, & x \in R-[0,1) \\
0, & 0 \leq x<1
\end{array}\right.
\end{aligned}
\)
\(
\begin{aligned}
&\text { At } x=0 \\
&\qquad \mathrm{RHL}=\lim _{x \rightarrow 0^{+}} 0=0
\end{aligned}
\)
\(
\text { and } \begin{aligned}
\mathrm{LHL} &=\lim _{x \rightarrow 0^{-}} \frac{\sin [x]}{[x]} \\
&=\lim _{h \rightarrow 0} \frac{\sin [0-h]}{[0-h]} \\
&=\lim _{h \rightarrow 0} \frac{\sin (-1)}{-1}=\sin 1
\end{aligned}
\)
Since, \(\quad \mathrm{RHL} \neq \mathrm{LHL}\)
\(\therefore\) Limit does not exist.
\(
\lim _{x \rightarrow 0} \frac{\sin |x|}{x} \text { is }
\)
(c)
\(\lim _{x \rightarrow 0} \frac{\sin |x|}{x}=?\)
L.H.L \(x \rightarrow 0-h[latex] where [latex]h>0\)
\(\lim _{h \rightarrow 0} \frac{\sin |0-h|}{0-h}=\lim _{h \rightarrow 0} \frac{\sin |-h|}{-h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=-1\)
R.H.L \(\quad x \rightarrow 0+h[latex] where [latex]h>0\)
\(\lim _{h \rightarrow 0} \frac{\sin |0+h|}{0+h}=\lim _{h \rightarrow 0} \frac{\sin |h|}{h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=1\)
\(L H L \neq R H L\)
\(\)\therefore\(\) The limit doesn’t exist.
Let \(f(x)=\begin{aligned}&x^2-1,0<x<2 \\&2 x+3,2 \leq x<3\end{aligned}\), the quadratic equation whose roots are \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) is
(d)Â
\(
\begin{aligned}
&\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-} x^2-1=3 \\
&\lim _{x \rightarrow 2+} f(x)=\lim _{x \rightarrow 2+} 2 x+3=7
\end{aligned}
\)
Then equation whose roots are 3,7
\(
\begin{aligned}
&x^2-(3+7) x+(3 \times 7)=0 \\
&\Rightarrow x^2-10 x+21=0
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x} \text { is }
\)
(b)
\(
\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0}\left\{\frac{\frac{2 \tan 2 x}{2 x}-1}{3-\frac{\sin x}{x}}\right\}=\frac{1}{2}
\)
Alternate solution: Apply L’Hospital’s rule,
\(
\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{2 \sec ^2 2 x-1}{3-\cos x}=\frac{2-1}{3-1}=\frac{1}{2}
\)
\(
\text { Let } f(x)=x-[x] ; \in \mathbf{R} \text {, then } f^{\prime}Â \frac{1}{2} \text { is }
\)
(b)
Let \(f(x)=x-[x] ; \in R\), then \(f^{\prime}\left(\frac{1}{2}\right)\) is 1
Explanation:
Given \(f(x)=x-[x]\)
We have to first check for differentiability of \(f(x)\) at \(x=\frac{1}{2}\)
\(
\begin{aligned}
&\therefore \text { Lf }^{\prime}\left(\frac{1}{2}\right)=\text { L.H.D } \\
&=\lim _{h \rightarrow 0} \frac{f\left[\frac{1}{2}-h\right]-f\left[\frac{1}{2}\right]}{-h} \\
&=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}-h\right)-\left[\frac{1}{2}-h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{-h} \\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{2}-h-0-\frac{1}{2}+0}{-h} \\
&=\frac{-h}{-h} \\
&=1
\end{aligned}
\)
\(
\begin{aligned}
&\text { Rf’ }\left(\frac{1}{2}\right)=\text { R.H.D } \\
&=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h} \\
&=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}+h\right)-\left[\frac{1}{2}+h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{h} \\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{2}+h-1-\frac{1}{2}+1}{h} \\
&=\frac{h}{h} \\
&=1 \\
&\text { Since L.H.D = R.H.D } \\
&\therefore f^{\prime}\left(\frac{1}{2}\right)=1
\end{aligned}
\)
If \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\), then \(\frac{d y}{d x}\) at \(x=1\) is
(d)
We have,
\(
y=\sqrt{x}+\frac{1}{\sqrt{x}}
\)
On taking differentiating both sides w.r.t x, we get \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2(x)^{\frac{3}{2}}}\)
Since, \(x=1\)
Therefore,
\(
\begin{aligned}
&\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=\frac{1}{2 \sqrt{1}}-\frac{1}{2(1)^{\frac{3}{2}}} \\
&\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=\frac{1}{2}-\frac{1}{2} \\
&\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=0
\end{aligned}
\)
\(
\text { If } f(x)=\frac{x-4}{2 \sqrt{x}}, \text { then } f^{\prime}(1) \text { is }
\)
(a)
As, \(f(x)=\frac{x-4}{2 \sqrt{x}}=\frac{1}{2}\left\{\frac{x}{\sqrt{x}}-\frac{4}{\sqrt{x}}\right\}=\frac{1}{2}\left\{x^{\frac{1}{2}}-4 x^{-\frac{1}{2}}\right\}\)
We know that,
\(
\begin{aligned}
\frac{d}{d x}\left(x^n\right) &=n x^{n-1} \\
\therefore f^{\prime}(x) &=\frac{d}{d x}(f(x))=\frac{1}{2} \frac{d}{d x}\left(x^{\frac{1}{2}}-4 x^{-\frac{1}{2}}\right) \\
\Rightarrow f^{\prime}(x) &=\frac{1}{2}\left\{\frac{d}{d x}\left(x^{\frac{1}{2}}\right)-4 \frac{d}{d x}\left(x^{-\frac{1}{2}}\right)\right\} \\
&=\frac{1}{2}\left\{\frac{1}{2} x^{-\frac{1}{2}}-4\left(-\frac{1}{2}\right) x^{-\frac{1}{2}-1}\right\} \\
& \Rightarrow f^{\prime}(x)=\frac{1}{2}\left\{\frac{1}{2} x^{-\frac{1}{2}}+2 x^{-\frac{3}{2}}\right\} \\
\therefore f^{\prime}(1) &=\frac{1}{2}\left\{\frac{1}{2}(1)^{-\frac{1}{2}}+2(1)^{-\frac{3}{2}}\right\}=\frac{1}{2}\left\{\frac{1}{2}+2\right\}=\frac{5}{4} \\
\therefore f^{\prime}(1) &=5 / 4
\end{aligned}
\)
If \(y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}\), then \(\frac{d y}{d x}\) is
(a)
\(
\begin{aligned}
&\mathrm{y}=\frac{1+\frac{1}{\mathrm{x}^2}}{1-\frac{1}{\mathrm{x}^2}} \\
&\mathrm{y}=\frac{\mathrm{x}^2+1}{\mathrm{x}^2-1} \\
&\therefore \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{\left(\mathrm{x}^2-1\right) \cdot 2 \mathrm{x}-\left(\mathrm{x}^2+1\right) \cdot 2 \mathrm{x}}{\left(\mathrm{x}^2-1\right)^2} \\
&=\frac{2 \mathrm{x}\left(\mathrm{x}^2-1-\mathrm{x}^2-1\right)}{\left(\mathrm{x}^2-1\right)^2} \\
&=\frac{2 \mathrm{x}(-2)}{\left(\mathrm{x}^2-1\right)^2} \\
&=\frac{-4 \mathrm{x}}{\left(\mathrm{x}^2-1\right)^2}
\end{aligned}
\)
\(
\text { If } y=\frac{\sin x+\cos x}{\sin x-\cos x} \text {, then } \frac{d y}{d x} \text { at } x=0 \text { is }
\)
The derivative of \(y\) will thus look like this
\(
\frac{d}{d x}(y)=\frac{(\cos x-\sin x) \cdot(\sin x-\cos x)-(\sin x+\cos x) \cdot(\cos x+\sin x)}{(\sin x-\cos x)^2}
\)
I’ll break this into two fractions to make the calculations easier to read
\(
\frac{d}{d x}(y)=y^{\prime}=\underbrace{\frac{f^{\prime}(x) \cdot g(x)}{g(x)^2}}_A-\underbrace{\frac{f(x) \cdot g^{\prime}(x)}{g(x)^2}}_B=A-B
\)
The first fraction will be equal to
\([
A=\frac{(\cos x-\sin x) \cdot(\sin x-\cos x)}{(\sin x-\cos x)^2}
\)
\(
A=\frac{\cos x \cdot \sin x-\cos ^2 x-\sin ^2 x+\sin x \cos x}{(\sin x-\cos x)^2}
\)
\(
A=\frac{2 \sin x \cos x-\left(\sin ^2 x+\cos ^2 x\right)}{(\sin x-\cos x)^2}
\)
You can further simplify this by using the fact that
\(
\begin{aligned}
&\sin ^2 x+\cos ^2 x=1 \\
&A=\frac{2 \sin x \cos x-1}{(\sin x-\cos x)^2}
\end{aligned}
\)
The second fraction will be equal to
\(
B=\frac{(\sin x+\cos x) \cdot(\cos x+\sin x)}{(\sin x-\cos x)^2}
\)
\(
B=\frac{(\sin x+\cos x)^2}{(\sin x-\cos x)^2}
\)
You can simplify this by using
\(
(a+b)^2=a^2+2 a b+b^2
\)
to get
\(
B=\frac{\sin ^2 x+2 \sin x \cos x+\cos ^2 x}{(\sin x-\cos x)^2}
\)
\(
B=\frac{2 \sin x \cos x+1}{(\sin x-\cos x)^2}
\)
Put the two fractions back together to get
\(
y^{\prime}=\frac{2 \sin x \cos x-1-(2 \sin x \cos x+1)}{(\sin x-\cos x)^2}
\)
\(
y^{\prime}=\frac{2 \sin x \cos x-1-2 \sin x \cos x-1}{(\sin x-\cos x)^2}
\)
Finally, you get
\(
y^{\prime}=-\frac{2}{(\sin x-\cos x)^2}
\)
If we put x=0, we get \(y^{\prime}=-2\)
\(
\text { If } y=\frac{\sin (x+9)}{\cos x} \text { then } \frac{d y}{d x} \text { at } x=0 \text { is }
\)
(a)
\(
\begin{aligned}
&\text { Given } y=\frac{\sin (x+9)}{\cos x} \\
&\frac{d y}{d x}=\frac{\cos x \cdot \cos (x+9)-\sin (x+9)(-\sin x)}{\cos ^2 x} \\
&=\frac{\cos x \cos (x+9)+\sin x \sin (x+9)}{\cos ^x} \\
&=\frac{\cos (x+9-x)}{\cos ^2 x} \\
&=\frac{\cos 9}{\cos 2 x} \\
&\therefore\left(\frac{d y}{d x}\right)_{\text {at } x=0}=\frac{\cos ^2 9}{\cos ^2 0} \\
&=\frac{\cos 9}{(1)^2} \\
&=\cos 9
\end{aligned}
\)
\(
\text { If } f(x)=1+x+\frac{x^2}{2}+\ldots+\frac{x^{100}}{100} \text {, then } f^{\prime}(1) \text { is equal to }
\)
(b)
\(
f(x)=1+x+\frac{x^2}{2}+\ldots+\frac{x^{100}}{100}
\)
Differentiating both sides with respect to \(x\), we get
\(
\begin{aligned}
&f^{\prime}(x)=\frac{d}{d x}\left(1+x+\frac{x^2}{2}+\ldots+\frac{x^{100}}{100}\right) \\
&=\frac{d}{d x}(1)+\frac{d}{d x}(x)+\frac{d}{d x}\left(\frac{x^2}{2}\right)+\ldots+\frac{d}{d x}\left(\frac{x^{100}}{100}\right) \\
&=\frac{d}{d x}(1)+\frac{d}{d x}(x)+\frac{1 d}{2 d x}\left(x^2\right)+\ldots+\frac{1 d}{100 d x}\left(x^{100}\right) \\
&=0+1+\frac{1}{2} \times 2 x+\ldots+\frac{1}{100} \times 100 x^{99} \\
&=1+x+x^2+\ldots+x^{99}
\end{aligned}
\)
Putting \(x=1\), we get
\(
\begin{aligned}
&f^{\prime}(1)=1+1+1+\ldots+1(100 \text { terms }) \\
&=100
\end{aligned}
\)
\(
\text { If } f(x)=\frac{x^n-a^n}{x-a} \text { for some constant ‘ } a \text { ‘, then } f^{\prime}(a) \text { is }
\)
(c)
Given \(f(x)=\frac{x^n-a^n}{x-a}\)
\(
f^{\prime}(x)=\frac{(x-a)\left(n \cdot x^{n-1}\right)-\left(x^n-a^n\right) \cdot 1}{(x-a)^2}
\)
\(
\therefore f^{\prime}(a)=\frac{(a-a)\left(n \cdot a^{n-1}\right)-\left(a^n-a^n\right)}{(a-a)^2}
\)
So \(f^{\prime}(a)=\frac{0}{0}\)
\(=\) does not exist
\(
\text { If } f(x)=x^{100}+x^{99}+\ldots+x+1 \text {, then } f^{\prime}(1) \text { is equal to }
\)
(a)
Given, \(f(x)=x^{100}+x^{99}+\ldots+x+1\)
\(\therefore f^{\prime}(\mathrm{x})=100 x^{99}+99 x^{98}+\ldots+1\)
So, \(f^{\prime}(1)=100+99+98+\ldots+1\)
\(=\frac{100}{2}[2 \times 100+(100-1)(-1)]\)
\(=50[200-99]\)
\(=50 \times 101\)
\(=5050\)
\(
\text { If } f(x)=1-x+x^2-x^3 \ldots-x^{99}+x^{100} \text {, then } f^{\prime}(1) \text { is equal to }
\)
(d) Given that \(f(x)=1-x+x^2-x^3+\ldots-x^{99}+x^{100}\)
\(
\begin{aligned}
&f^{\prime}(x)=-1+2 x-3 x^2+\ldots-99 x^{98}+100 x^{99} \\
&\therefore f^{\prime}(1)=-1+2+3+\ldots-99+100 \\
&=(-1-3-5 \ldots-99)+(2+4+6+\ldots+100) \\
&=\frac{50}{2}[2 \times-1+(50-1)(-2)]+\frac{50}{2}[2 \times 2-(50-1) 2] \\
&=25[-2-98]+25[4+98] \\
&=25 \times-100+25 \times 102 \\
&=25[-100+102] \\
&=25 \times 2 \\
&=50
\end{aligned}
\)
\(
\text { If } f(x)=\frac{\tan x}{x-\pi} \text {, then } \lim _{x \rightarrow \pi} f(x)= ?
\)
(a) Consider the given function.
\(
f(x)=\frac{\tan x}{x-\pi}
\)
Since, \(\lim _{x \rightarrow \pi} f(x)\)
Thus,
\(\lim _{x \rightarrow \pi}\left(\frac{\tan x}{x-\pi}\right)\)
This is the \(\frac{0}{0}\) form.
So, apply L-Hospital rule,
\(\lim _{x \rightarrow \pi}\left(\frac{\sec ^2 x}{1-0}\right)\)
\(\lim _{x \rightarrow \pi}\left(\sec ^2 x\right)\)
\(=(\sec \pi)^2\)
\(=(-1)^2\)
\(=1\)
\(
\lim _{x \rightarrow 0} \sin m x \cot \frac{x}{\sqrt{3}}=2, \text { then } m=?
\)
(a)
\(\lim _{x \rightarrow 0} \sin m x \cot \frac{x}{\sqrt{3}}=2\) \(\Rightarrow \lim _{x \rightarrow 0} \sin m x \frac{\cos \frac{x}{\sqrt{3}}}{\sin \frac{x}{\sqrt{3}}}; \quad \frac{0}{0}\) from
\(\Rightarrow \lim _{x \rightarrow 0} \frac{m \cos m x \cos \frac{x}{\sqrt{3}}-\frac{1}{\sqrt{3}} \sin \frac{x}{\sqrt{3}} \sin m x}{\frac{1}{\sqrt{3}} \cos \frac{x}{\sqrt{3}}}=2\) \(\Rightarrow \frac{\mathrm{m}}{\left(\frac{1}{\sqrt{3}}\right)}=2\)
\(
\Rightarrow \mathrm{m}=\frac{2}{\sqrt{3}}
\)
\(
\text { if } y=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots, \text { then } \frac{d y}{d x}=?
\)
(c)
As, \(y=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\)
We know that,
\(
\begin{aligned}
&\frac{d}{d x}\left(x^n\right)=n x^{n-1} \\
&\therefore \frac{d y}{d x}=\frac{d}{d x}\left(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right) \\
&\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(1)+\frac{d}{d x}\left(\frac{x}{1 !}\right)+\frac{d}{d x}\left(\frac{x^2}{2 !}\right)+\cdots \\
&\Rightarrow \frac{d y}{d x}=0+1+\frac{2 x}{2 !}+\frac{3 x^2}{3 !}+\cdots \\
&\Rightarrow \frac{d y}{d x}=1+\frac{x}{1}+\frac{x^2}{2 !}+\cdots
\end{aligned}
\)
Clearly, in comparison with y, we can say that-
\(
\frac{d y}{d x}=y
\)
\(
\lim _{x \rightarrow 3^{+}} \frac{x}{[x]}=?
\)
(c) Given, \(\lim _{x \rightarrow 3^{+}} \frac{x}{[x]}=\lim _{h \rightarrow 0} \frac{(3+h)}{[3+h]}\) = \(\lim _{h \rightarrow 0} \frac{3+h}{3}=1\)
\(
\text { Evaluate } \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}Â
\)
Given that \(\operatorname{Lim}_{x \rightarrow 3} \frac{x^2-9}{x-3}=\operatorname{Lim}_{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)}=\operatorname{Lim}_{x \rightarrow 3} x+3\)
Taking limit, we have
\(
3+3=6
\)
Hence, the answer is 6.
\(
\text { Evaluate } \lim _{x \rightarrow \frac{1}{2}} \frac{4 x^2-1}{2 x-1}
\)
\(
\text { Given that: } \operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{4 x^2-1}{2 x-1}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{(2 x)^2-(1)^2}{2 x-1}=\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{(2 x+1)(2 x-1)}{2 x-1} \\
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}}(2 x+1)
\end{aligned}
\)
Taking limit, we have
\(
=2 \times \frac{1}{2}+1=1+1=2
\)
Hence, the answer is 2.
\(
\text { Evaluate } \lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}
\)
Given that \(\operatorname{Lim}_{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\)
\(
=\operatorname{Lim}_{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h[\sqrt{x+h}+\sqrt{x}]} \times \sqrt{x+h}+\sqrt{x}
\)
[Rationalizing the denominator]
\(
=\operatorname{Lim}_{h \rightarrow 0} \frac{x+h-x}{h[\sqrt{x+h}+\sqrt{x}]}
\)
\(
=\operatorname{Lim}_{h \rightarrow 0} \frac{h}{h[\sqrt{x+h}+\sqrt{x}]}=\operatorname{Lim}_{h \rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}
\)
Taking the limits, we have
\(
\frac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2 \sqrt{x}}
\)
Hence, the answer is \(\frac{1}{2 \sqrt{x}}\).
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{(x+2)^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x}\)
Put \(x+2=y \quad \Rightarrow \quad x=y-2\)
\(
\begin{aligned}
& =\operatorname{Lim}_{y-2 \rightarrow 0} \frac{y^{1 / 3}-2^{1 / 3}}{y-2}=\operatorname{Lim}_{y \rightarrow 2} \frac{y^{1 / 3}-2^{1 / 3}}{y-2} \\
& =\frac{1}{3} \cdot(2)^{\frac{1}{3}-1}=\frac{1}{3} \cdot 2^{-2 / 3} \quad\left[\text { using } \operatorname{Lim}_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right]
\end{aligned}
\)
Hence the answer is \(\frac{1}{3}(2)^{-2 / 3}\)
\(
\text { Evaluate } \lim _{x \rightarrow 1} \frac{(1+x)^6-1}{(1+x)^2-1}
\)
Given that: \(\operatorname{Lim}_{x \rightarrow 0} \frac{(1+x)^6-1}{(1+x)^2-1}[latex]
Dividing the numerator and denominator by [latex]x[latex], we get
[latex]
=\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{(1+x)^6-1}{x}}{\frac{(1+x)^2-1}{x}}
\)
\(
\text { Putting } 1+x=y \Rightarrow x=y-1
\)
\(
=\operatorname{Lim}_{\substack{y \rightarrow 1 \rightarrow 0 \\ \therefore y \rightarrow 1}} \frac{\frac{y^6-(1)^6}{y-1}}{\frac{y^2-(1)^2}{y-1}}=\frac{\operatorname{Lim}_{y \rightarrow 1} \frac{y^6-(1)^6}{y-1}}{\operatorname{Lim}_{y \rightarrow 1} \frac{y^2-(1)^2}{y-1}} \left[\operatorname{Lim}_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\operatorname{Lim}_{x \rightarrow a} f(x)}{\operatorname{Lim}_{x \rightarrow a} g(x)}\right]
\)
\(
=\frac{6 \cdot(1)^{6-1}}{2 \cdot(1)^{2-1}}=\frac{6}{2}=3 \quad\left[\text { using } \operatorname{Lim}_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right]
\)
Hence, the required answer is 3.
\(
\text { Evaluate } \lim _{x \rightarrow a} \frac{(2+x)^{\frac{5}{2}}-(a+2)^{\frac{5}{2}}}{x-a}
\)
Given that: \(\operatorname{Lim}_{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{(2+x)-(a+2)}\)
\(
=\operatorname{Lim}_{2+x \rightarrow a+2} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{(2+x)-(a+2)}
\)
\(
=\frac{5}{2}(a+2)^{5 / 2-1}=\frac{5}{2}(a+2)^{3 / 2}\left[\because \operatorname{Lim}_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right]
\)
Hence, the required answer is \(\frac{5}{2}(a+2)^{3 / 2}\).
\(
\text { Evaluate } \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}
\)
\(
\text { Given that } \operatorname{Lim}_{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}
\)
\(
=\operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{x}\left[(x)^{7 / 2}-1\right]}{\sqrt{x}-1}=\operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{x} \frac{\left[x^{7 / 2}-(1)^{7 / 2}\right]}{x-1}}{\frac{(x)^{1 / 2}-(1)^{1 / 2}}{x-1}}
\)
[Dividing the numerator and denominator of \(x-1\) ]
\(
\begin{aligned}
\operatorname{Lim}_{x \rightarrow 1} \frac{\frac{(x)^{7 / 2}-(1)^{7 / 2}}{x-1}}{\frac{(x)^{1 / 2}-(1)^{1 / 2}}{x-1}} \times \operatorname{Lim}_{x \rightarrow 1} \sqrt{x} \\
{\left[\because \operatorname{Lim}_{x \rightarrow a} f(x) g(x)=\operatorname{Lim}_{x \rightarrow a} f(x) \cdot \operatorname{Lim}_{x \rightarrow a} g(x)\right] }
\end{aligned}
\)
\(
=\frac{\frac{7}{2}(1)^{7 / 2-1}}{\frac{1}{2}(1)^{1 / 2-1}} \times \sqrt{1}=\frac{7 / 2}{1 / 2}=7
\)
Hence the required answer is 7.
\(
\text { Evaluate } \lim _{x \rightarrow 2} \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 2} \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}}\)
Rationalizing the denominator, we get
\(
=\operatorname{Lim}_{x \rightarrow 2} \frac{(x-2)(x+2)[\sqrt{3 x-2}+\sqrt{x+2}]}{[\sqrt{3 x-2}-\sqrt{x+2}][\sqrt{3 x-2}+\sqrt{x+2}]}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 2} \frac{(x-2)(x+2)[\sqrt{3 x-2}+\sqrt{x+2}]}{3 x-2-x-2} \\
& =\operatorname{Lim}_{x \rightarrow 2} \frac{(x-2)(x+2)[\sqrt{(3 x-2)}+\sqrt{x+2}]}{2 x-4} \\
& =\operatorname{Lim}_{x \rightarrow 2} \frac{(x-2)(x+2)[\sqrt{(3 x-2)}+\sqrt{x+2}]}{2(x-2)} \\
& =\operatorname{Lim}_{x \rightarrow 2} \frac{(x+2)[\sqrt{3 x-2}+\sqrt{x+2}]}{2}
\end{aligned}
\)
Taking limits, we have
\(
=\frac{(2+2)[\sqrt{6-2}+\sqrt{2+2}]}{2}=\frac{4[2+2]}{2}=\frac{4 \times 4}{2}=8
\)
Hence, the required answer is 8.
\(
\text { Evaluate } \lim _{x \rightarrow \sqrt{2}} \frac{x^4-4}{x^2+3 \sqrt{2 x}-8}
\)
Given that \(\operatorname{Lim}_{x \rightarrow \sqrt{2}} \frac{x^4-4}{x^2+3 \sqrt{2} x-8}\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow \sqrt{2}} \frac{\left(x^2-2\right)\left(x^2+2\right)}{x^2+4 \sqrt{2} x-\sqrt{2} x-8} \\
& =\operatorname{Lim}_{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})(x-\sqrt{2})\left(x^2+2\right)}{x(x+4 \sqrt{2})-\sqrt{2}(x+4 \sqrt{2})} \\
& =\operatorname{Lim}_{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})(x-\sqrt{2})\left(x^2+2\right)}{(x+4 \sqrt{2})(x-\sqrt{2})}=\operatorname{Lim}_{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})\left(x^2+2\right)}{x+4 \sqrt{2}}
\end{aligned}
\)
Taking limits we have
\(
=\frac{(\sqrt{2}+\sqrt{2})(2+2)}{\sqrt{2}+4 \sqrt{2}}=\frac{2 \sqrt{2} \times 4}{5 \sqrt{2}}=\frac{8}{5}
\)
Hence, the required answer is \(\frac{8}{5}\).
\(
\text { Evaluate } \lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2}
\)
Given that
\(
=\operatorname{Lim}_{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2} \left[\frac{0}{0} \text { form }\right] \\
\)
\(
=\operatorname{Lim}_{x \rightarrow 1} \frac{x^7-x^5-x^5+1}{x^3-x^2-2 x^2+2}=\operatorname{Lim}_{x \rightarrow 1} \frac{x^5\left(x^2-1\right)-1\left(x^5-1\right)}{x^2(x-1)-2\left(x^2-1\right)}
\)
Dividing the numerator and denominator by \((x-1)[latex] we get
[latex]
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 1} \frac{x^5\left(\frac{x^2-1}{x-1}\right)-1\left(\frac{x^5-1}{x-1}\right)}{x^2\left(\frac{x-1}{x-1}\right)-2\left(\frac{x^2-1}{x-1}\right)} \\
& =\frac{\operatorname{Lim}_{x \rightarrow 1} x^5(x+1)-\operatorname{Lim}\left(\frac{x^5-(1)^5}{x-1}\right)}{\operatorname{Lim}_{x \rightarrow 1} x^2-2 \operatorname{Lim}_{x \rightarrow 1}(x+1)} \\
& =\frac{1(2)-5 \cdot(1)^{5-1}}{1-2(2)}=\frac{2-5}{1-4}=\frac{-3}{-3}=1
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}\)
\(
=\operatorname{Lim}_{x \rightarrow 0} \frac{\left[\sqrt{1+x^3}-\sqrt{1-x^3}\right]\left[\sqrt{1+x^3}+\sqrt{1-x^3}\right]}{x^2\left[\sqrt{1+x^3}+\sqrt{1-x^3}\right]}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-x^3\right)}{x^2\left[\sqrt{1+x^3}+\sqrt{1-x^3}\right]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{1+x^3-1+x^3}{x^2\left[\sqrt{1+x^3}+\sqrt{1-x^3}\right]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 x^3}{x^2\left[\sqrt{1+x^3}+\sqrt{1-x^3}\right]}=\operatorname{Lim}_{x \rightarrow 0} \frac{2 x}{\sqrt{1+x^3}+\sqrt{1-x^3}}=0
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow-3} \frac{x^3+27}{x^5+243}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 3} \frac{x^3+27}{x^5+243}\) \(=\operatorname{Lim}_{x \rightarrow 3} \frac{\frac{x^3+(3)^3}{x-3}}{\frac{x^5+(3)^5}{x-3}} \quad\) [Dividing the \(Nr\) and Den. by \(x-3\) ]
\(
\begin{aligned}
& =\frac{\operatorname{Lim}_{x \rightarrow 3}\left(\frac{x^3-(-3)^3}{x+3}\right)}{\operatorname{Lim}_{x \rightarrow 3}\left(\frac{x^5-(-3)^5}{x+3}\right)} \quad\left[\operatorname{Lim}_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\operatorname{Lim}_{x \rightarrow a} f(x)}{\operatorname{Lim}_{x \rightarrow a} g(x)}\right] \\
& =\frac{3(-3)^{3-1}}{5(-3)^{5-1}}=\frac{3 \times(-3)^2}{5 \times(-3)^4}=\frac{1}{5 \times 3}=\frac{1}{15}
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow \frac{1}{2}}\left(\frac{8 x-3}{2 x-1}-\frac{4 x^2+1}{4 x^2-1}\right)
\)
Given that \(\operatorname{Lim}_{x \rightarrow \frac{1}{2}}\left(\frac{8 x-3}{2 x-1}-\frac{4 x^2+1}{4 x^2-1}\right)\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}}\left[\frac{(8 x-3)(2 x+1)-\left(4 x^2+1\right)}{\left(4 x^2-1\right)}\right] \\
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}}\left[\frac{16 x^2-6 x+8 x-3-4 x^2-1}{4 x^2-1}\right] \\
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}}\left[\frac{12 x^2+2 x-4}{4 x^2-1}\right]=\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{2\left(6 x^2+x-2\right)}{4 x^2-1} \\
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{2\left[6 x^2+4 x-3 x-2\right]}{(2 x+1)(2 x-1)}=\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{2[2 x(3 x+2)-1(3 x+2)]}{(2 x+1)(2 x-1)} \\
& =\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{2(3 x+2)(2 x-1)}{(2 x+1)(2 x-1)}=\operatorname{Lim}_{x \rightarrow \frac{1}{2}} \frac{2(3 x+2)}{(2 x+1)}
\end{aligned}
\)
Taking limit, we have
\(
=\frac{2\left(3 \times \frac{1}{2}+2\right)}{2 \times \frac{1}{2}+1}=\frac{2\left(\frac{7}{2}\right)}{2}=\frac{7}{2}
\)
\(
\text { Find ‘ } n \text { ‘, if } \lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80, n \in N
\)
\(
\text { Given that } \operatorname{Lim}_{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80
\)
\(
\begin{aligned}
& =n \cdot(2)^{n-1}=80 \quad\left[\because \quad \operatorname{Lim}_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right] \\
& =n \times 2^{n-1}=5 \times(2)^{5-1} \quad \\
n & =5
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow a} \frac{\sin 3 x}{\sin 7 x}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 0} \frac{\sin 3 x}{\sin 7 x}\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{\sin 7 x}{7 x} \times 7 x}=\frac{\operatorname{Lim}_{3 x \rightarrow 0}\left(\frac{\sin 3 x}{3 x}\right)}{\operatorname{Lim}_{7 x \rightarrow 0}\left(\frac{\sin 7 x}{7 x}\right)} \times \frac{3}{7} \\
& =\frac{1}{1} \times \frac{3}{7}=\frac{3}{7} \quad\left[\because \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\end{aligned}
\)
Hence, the required answer is \(\frac{3}{7}\).
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{\sin ^2 4 x}
\)
\(
\begin{aligned}
& \text { Given that } \operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2 2 x}{\sin ^2 4 x} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2 2 x}{\sin ^2 2(2 x)} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2 2 x}{4 \sin ^2 2 x \cdot \cos ^2 2 x} [\sin 2 x=2 \sin x \cos x] \\
&
\end{aligned}
\)
\(
=\frac{1}{4 \cos ^2 2 x}
\)
Taking limit we have
\(
=\frac{1}{4 \cdot \cos ^2 0}=\frac{1}{4}
\)
Hence, the required answer is \(\frac{1}{4}\).
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^2}
\)
Given that \(\operatorname{Lim}_{x \rightarrow 0} \frac{1-\cos 2 x}{x^2}\) \(=\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2} \left[\cos 2 x=1-2 \sin ^2 x\right]\)
\(
=\operatorname{Lim}_{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right)^2=2 \times 1=2 \quad\left[\because \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\)
Hence, the required answer is 2.
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}
\)
\(
\begin{aligned}
& \text { Given that } \operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^3}=\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^3} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin x}{x}\left(\frac{1-\cos x}{x^2}\right)=\operatorname{Lim}_{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right)\left(\frac{2 \sin ^2 x / 2}{x^2}\right)
\end{aligned}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right)\left(2 \frac{\sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{1}{4}\right) \\
& =\operatorname{Lim}_{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right) 2\left[\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2\right] \cdot \frac{1}{4} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{4}{4}\left(\frac{\sin x}{x}\right) \underset{\frac{x}{2} \rightarrow 0}{\operatorname{Lim}}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2 \\
& =1 \cdot 1 \cdot(1)^2=1 \left[\because \quad \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}
\)
\(
\text { Given that } \operatorname{Lim}_{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0}\left(\frac{2 \sin ^2 \frac{m}{2} x}{2 \sin ^2 \frac{n}{2} x}\right) \\
& =\operatorname{Lim}_{x \rightarrow 0}\left(\frac{\sin \frac{m}{2} x}{\sin \frac{n}{2} x}\right)^2
\end{aligned}
\)
\(
=\frac{\operatorname{Lim}_{x \rightarrow 0}\left(\frac{\sin \frac{m}{2} x}{\frac{m}{2} x} \times \frac{m}{2} x\right)^2}{\operatorname{Lim}_{x \rightarrow 0}\left(\frac{\sin \frac{n}{2} x}{\sin \frac{n}{2} x} \times \frac{n}{2} x\right)^2}=\frac{1 \cdot \frac{m^2}{4} x^2}{1 \cdot \frac{n^2}{4} x^2}\left[\because \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\)
Hence, the required answer is \(\frac{m^2}{n^2}\).
\(
\text { Evaluate } \lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}
\)
Given that \(\operatorname{Lim}_{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}\)
\(
=\operatorname{Lim}_{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{2 \sin ^2 3 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)} \left[\because 1-\cos \theta=2 \sin ^2 \theta / 2\right]
\)
\(
=\operatorname{Lim}_{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{2} \sin 3 x}{\sqrt{2}\left(\frac{\pi-3 x}{3}\right)}=\operatorname{Lim}_{\substack{x \rightarrow \pi \\ \therefore \pi-3 x \rightarrow 0}} \frac{3 \cdot \sin (\pi-3 x)}{\pi-3 x} =3 \left[\because \quad \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\)
\(
\text { Evaluate }\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}
\)
Given that
\(
\begin{aligned}
& \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}} \\
& =\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)}{x-\frac{\pi}{4}} \\
& =\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(\cos \frac{\pi}{4} \sin x-\sin \frac{\pi}{4} \cos x\right)}{x-\frac{\pi}{4}}
\end{aligned}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{\substack{x \rightarrow \frac{\pi}{4} \\
\therefore x-\frac{\pi}{4} \rightarrow 0}} \frac{\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}} \\
& \sqrt{2} \cdot 1=\sqrt{2} \\
&
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}
\)
\(
\begin{aligned}
\text { Given that } & \operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}} \\
= & \operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{2\left[\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right]}{x-\frac{\pi}{6}} \\
= & \operatorname{Lim}_{x \rightarrow \frac{\pi}{6}}=\frac{2\left[\cos \frac{\pi}{6} \sin x-\sin \frac{\pi}{6} \cos x\right]}{x-\frac{\pi}{6}}
\end{aligned}
\)
\(
=\operatorname{Lim}_{\substack{x \rightarrow \frac{\pi}{6}}} \frac{2 \sin \left(x-\frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)} \left[\because \quad \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\)
\(
\therefore x-\frac{\pi}{6} \rightarrow 0
\)
\(
=2 \cdot 1=2
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}
\)
\(
\begin{aligned}
& \text { Given that: } \operatorname{Lim}_{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\left(\frac{\sin 2 x+3 x}{2 x}\right) \times 2 x}{\left(\frac{2 x+\tan 3 x}{3 x}\right) \times 3 x}=\operatorname{Lim}_{x \rightarrow 0} \frac{\left(\frac{\sin 2 x}{2 x}+\frac{3 x}{2 x}\right) \times 2 x}{\left(\frac{2 x}{3 x}+\frac{\tan 3 x}{3 x}\right) \times 3 x} \\
& =\frac{\left(\operatorname{Lim}_{2 x \rightarrow 0} \frac{\sin 2 x}{2 x}+\frac{3}{2}\right)}{\left[\frac{2}{3}+\operatorname{Lim}_{3 x \rightarrow 0} \frac{\tan 3 x}{3 x}\right]} \times \frac{2}{3} \quad\left[\because \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\
&
\end{aligned}
\)
\(
\begin{array}{ll}
=\left(\frac{1+\frac{3}{2}}{\frac{2}{3}+1}\right) \times \frac{2}{3} & {\left[\because \operatorname{Lim}_{x \rightarrow 0} \frac{\tan x}{x}=1\right]} \\
=\frac{5 / 2}{5 / 3} \times \frac{2}{3}=\frac{3}{2} \times \frac{2}{3}=1 &
\end{array}
\)
\(
\text { Evaluate } \lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}
\)
\(
\begin{aligned}
\text { Given that: } & \operatorname{Lim}_{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}} \\
= & \operatorname{Lim}_{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} \\
= & \operatorname{Lim}_{x \rightarrow a} \frac{(\sin x-\sin a)(\sqrt{x}+\sqrt{a})}{x-a}
\end{aligned}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow a} \frac{\left(2 \cos \frac{x+a}{2} \cdot \sin \frac{x-a}{2}\right)(\sqrt{x}+\sqrt{a})}{x-a} \\
& =\operatorname{Lim}_{\substack{x \rightarrow a \\
x-a} 0}\left(2 \cos \frac{x+a}{2} \cdot \frac{\sin \frac{x-a}{2}}{2 \times \frac{x-a}{2}}\right)(\sqrt{x}+\sqrt{a})
\end{aligned}
\)
\(
=\operatorname{Lim}_{x \rightarrow a} \cos \left(\frac{x+a}{2}\right)(\sqrt{x}+\sqrt{a}) \quad\left[\because \operatorname{Lim}_{\frac{x-a}{2} \rightarrow 0} \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}}=1\right]
\)
Taking limit we have
\(
=\cos \left(\frac{a+a}{2}\right)(\sqrt{a}+\sqrt{a})=\cos a \times 2 \sqrt{a}=2 \sqrt{a} \cdot \cos a
\)
\(
\text { Evaluate } \lim _{x \rightarrow \frac{\pi}{6}} \frac{\cot ^2 x-3}{\operatorname{cosec} x-2}
\)
\(
\begin{aligned}
& \text { Given that } \operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{\cot ^2 x-3}{\operatorname{cosec} x-2} \\
& \quad=\operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{\operatorname{cosec}^2 x-1-3}{\operatorname{cosec} x-2}=\operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{\operatorname{cosec}^2 x-4}{\operatorname{cosec} x-2}
\end{aligned}
\)
\(
=\operatorname{Lim}_{x \rightarrow \frac{\pi}{6}} \frac{(\operatorname{cosec} x+2)(\operatorname{cosec} x-2)}{(\operatorname{cosec} x-2)}=\operatorname{Lim}_{x \rightarrow \frac{\pi}{6}}(\operatorname{cosec} x+2)
\)
Taking limit we have
\(
=\operatorname{cosec} \frac{\pi}{6}+2=2+2=4
\)
\(\)
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}
[/latex
Given that \(\operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x} \times \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2-(1+\cos x)}{\sin ^2 x[\sqrt{2}+\sqrt{1+\cos x}]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{1-\cos x}{\sin ^2 x[\sqrt{2}+\sqrt{1+\cos x}]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2 x / 2}{(2 \sin x / 2 \cos x / 2)^2} \times \overline{[\sqrt{2}+\sqrt{1+\cos x}]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2 x / 2}{4 \sin ^2 x / 2 \cos ^2 x / 2} \times \frac{1}{[\sqrt{2}+\sqrt{1+\cos x}]} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2}{4 \cos ^2 \frac{x}{2}} \times \frac{1}{[\sqrt{2}+\sqrt{1+\cos x}]}
\end{aligned}
\)
Taking limit, we get
\(
=\frac{2}{4 \cos ^2 0} \times \frac{1}{(\sqrt{2}+\sqrt{2})}=\frac{1}{2} \times \frac{1}{2 \sqrt{2}}=\frac{1}{4 \sqrt{2}}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}
\)
Given that: \(\operatorname{Lim}_{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}\)
\(
=\operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}-\frac{2 \sin 3 x}{x}+\frac{\sin 5 x}{x}
\)
\(
=\operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}-\operatorname{Lim}_{3 x \rightarrow 0} 2\left(\frac{\sin 3 x}{3 x}\right) \times 3+\operatorname{Lim}_{5 x \rightarrow 0}\left(\frac{\sin 5 x}{5 x}\right) \times 5
\)
\(
=1-6+5=0
\)
Hence, the required answer is 0 .
\(
\text { If } \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2} \text {, then find the value of } k \text {. }
\)
Given that \(\lim _{x \rightarrow 1}\left(\frac{x^4-1}{x-1}\right)=4(1)^{4-1}=4\)
\(
\begin{aligned}
& \lim _{x \rightarrow k}\left(\frac{x^3-k^3}{x^2-k^2}\right)=\lim _{x \rightarrow k}\left(\frac{(x-k)\left(x^2+k^2+x k\right)}{(x-k)(x+k)}\right)=\lim _{x \rightarrow k}\left(\frac{x^2+k^2+x k}{x+k}\right) \\
& =\frac{3 k}{2} \\
& \quad \frac{3 k}{2}=4 \\
& k=8 / 3
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\frac{x^4+x^3+x^2+1}{x}
\)
\(
\begin{aligned}
\frac{d}{d x}\left(\frac{x^4+x^3+x^2+1}{x}\right) & =\frac{d}{d x}\left(x^3+x^2+x+\frac{1}{x}\right) \\
& =3 x^2+2 x+1-\frac{1}{x^2}
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\left(x+\frac{1}{x}\right)^3
\)
\(
\begin{aligned}
& \frac{d}{d x}\left(x+\frac{1}{x}\right)^3=\frac{d}{d x}\left(x^3+\frac{1}{x^3}+3 x+\frac{3}{x}\right) \\
&=\frac{d}{d x}\left(x^3+x^{-3}+3 x+3 \cdot x^{-1}\right)=3 x^2-3 x^{-4}+3-3 \cdot x^{-2} \\
&=3 x^2-\frac{3}{x^4}+3-\frac{3}{x^2}
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\((3 x+5)(1+\tan x)\)
\(
\frac{d}{d x}(3 x+5)(1+\tan x)
\)
\(
\begin{aligned}
& =(3 x+5) \frac{d}{d x}(1+\tan x)+(1+\tan x) \frac{d}{d x}(3 x+5) \\
& =(3 x+5)\left(\sec ^2 x\right)+(1+\tan x)(3) \\
& =3 x \sec ^2 x+5 \sec ^2 x+3+3 \tan x \quad \text { [using product rule] }
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
(\sec x-1)(\sec x+1)
\)
\(
\begin{aligned}
& \frac{d}{d x}(\sec x-1)(\sec x+1) \\
& =(\sec x-1) \cdot \frac{d}{d x}(\sec x+1)+(\sec x+1) \frac{d}{d x}(\sec x-1) \\
& \text { [using product rule] } \\
& =(\sec x-1)(\sec x \tan x)+(\sec x+1)(\sec x \tan x) \\
& =\sec x \tan x(\sec x-1+\sec x+1) \\
& =\sec x \tan x \cdot 2 \sec x=2 \sec ^2 x \cdot \tan x \\
&
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\frac{3 x+4}{5 x^2-7 x+9}
\)
\(
\begin{aligned}
& \frac{d}{d x}\left(\frac{3 x+4}{5 x^2-7 x+9}\right) \\
& \quad=\frac{\left(5 x^2-7 x+9\right) \frac{d}{d x}(3 x+4)-(3 x+4) \cdot \frac{d}{d x}\left(5 x^2-7 x+9\right)}{\left(5 x^2-7 x+9\right)^2}
\end{aligned}
\)
[Using quotient rule]
\(
\begin{aligned}
& =\frac{\left(5 x^2-7 x+9\right)(3)-(3 x+4)(10 x-7)}{\left(5 x^2-7 x+9\right)^2} \\
& =\frac{15 x^2-21 x+27-30 x^2+21 x-40 x+28}{\left(5 x^2-7 x+9\right)^2} \\
& =\frac{-15 x^2-40 x+55}{\left(5 x^2-7 x+9\right)^2}=\frac{55-40 x-15 x^2}{\left(5 x^2-7 x+9\right)^2}
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\frac{x^5-\cos x}{\sin x}
\)
\(
\begin{gathered}
\frac{d}{d x}\left(\frac{x^5-\cos x}{\sin x}\right)=\frac{\sin x \cdot \frac{d}{d x}\left(x^5-\cos x\right)-\left(x^5-\cos x\right) \cdot \frac{d}{d x}(\sin x)}{\sin ^2 x} \text { [Using quotient rule] } \\
\quad=\frac{\sin x\left(5 x^4+\sin x\right)-\left(x^5-\cos x\right)(\cos x)}{\sin ^2 x}
\end{gathered}
\)
\(
\begin{aligned}
& =\frac{5 x^4 \cdot \sin x+\sin ^2 x-x^5 \cos x+\cos ^2 x}{\sin ^2 x} \\
& =\frac{5 x^4 \sin x-x^5 \cos x+\left(\sin ^2 x+\cos ^2 x\right)}{\sin ^2 x} \\
& =\frac{5 x^4 \sin x-x^5 \cos x+1}{\sin ^2 x}
\end{aligned}
\)
Hence, the required answer is \(\frac{5 x^4 \sin x-x^5 \cos x+1}{\sin ^2 x}\).
Differentiate each of the functions w. r. to \(x\) \(\frac{x^2 \cos \frac{\pi}{4}}{\sin x}\)
\(
\frac{d}{d x}\left(\frac{x^2 \cos \frac{\pi}{4}}{\sin x}\right)=\cos \frac{\pi}{4} \cdot \frac{d}{d x}\left(\frac{x^2}{\sin x}\right)
\)
\(
=\frac{\frac{1}{\sqrt{2}}\left[\sin x \cdot \frac{d}{d x}\left(x^2\right)-x^2 \cdot \frac{d}{d x}(\sin x)\right]}{\sin ^2 x} \text { [Using quotient rule] }
\)
\(
\begin{aligned}
& =\frac{1}{\sqrt{2}}\left[\frac{\sin x \cdot 2 x-x^2 \cos x}{\sin ^2 x}\right]=\frac{1}{\sqrt{2}}\left[\frac{2 x}{\sin x}-\frac{x^2 \cos x}{\sin ^2 x}\right] \\
& =\frac{1}{\sqrt{2}}\left[2 x \operatorname{cosec} x-x^2 \cot x \operatorname{cosec} x\right] \\
& =\frac{x}{\sqrt{2}} \operatorname{cosec} x[2-x \cot x]
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\left(a x^2+\cot x\right)(p+q \cos x)
\)
\(
\begin{aligned}
& \frac{d}{d x}\left(a x^2+\cot x\right)(p+q \cos x) \\
& \quad=\left(a x^2+\cot x\right) \frac{d}{d x}(p+q \cos x)+(p+q \cos x) \frac{d}{d x}\left(a x^2+\cot x\right) \text { [Using Product Rule] }
\end{aligned}
\)
\(
=\left(a x^2+\cot x\right)(-q \sin x)+(p+q \cos x)\left(2 a x-\operatorname{cosec}^2 x\right)
\)
Hence, the required answer is
\(
=\left(a x^2+\cot x\right)(-q \sin x)+(p+q \cos x)\left(2 a x-\operatorname{cosec}^2 x\right)
\)
Differentiate each of the functions w. r. to \(x\)
\(
\frac{a+b \sin x}{c+d \cos x}
\)
\(
\frac{d}{d x}\left(\frac{a+b \sin x}{c+d \cos x}\right)
\)
\(
\begin{aligned}
& =\frac{(c+d \cos x) \cdot \frac{d}{d x}(a+b \sin x)-(a+b \sin x) \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^2} \text { [Using quotient rule] } \\
& =\frac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^2} \\
& =\frac{c b \cos x+b d \cos ^2 x+a d \sin x+b d \sin ^2 x}{(c+d \cos x)^2} \\
& =\frac{c b \cos x+a d \sin x+b d\left(\cos ^2 x+\sin ^2 x\right)}{(c+d \cos x)^2} \\
& =\frac{c b \cos x+a d \sin x+b d}{(c+d \cos x)^2}
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
(\sin x+\cos x)^2
\)
\(
\begin{aligned}
& \frac{d}{d x}(\sin x+\cos x)^2=\frac{d}{d x}(\sin x+\cos x)(\sin x+\cos x) \\
& =(\sin x+\cos x) \frac{d}{d x}(\sin x+\cos x) \\
& +(\sin x+\cos x) \frac{d}{d x}(\sin x+\cos x) \\
& =2(\sin x+\cos x) \frac{d}{d x}(\sin x+\cos x) \\
& =2(\sin x+\cos x)(\cos x-\sin x)=2\left(\cos ^2 x-\sin ^2 x\right)=2 \cos 2 x \\
&
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
(2 x-7)^2(3 x+5)^3
\)
\(
\begin{aligned}
& \frac{d}{d x}(2 x-7)^2(3 x+5)^3 \\
& =(2 x-7)^2 \cdot \frac{d}{d x}(3 x+5)^3+(3 x+5)^3 \cdot \frac{d}{d x}(2 x-7)^2 \text { [Using product Rule] } \\
& =(2 x-7)^2 \cdot 3(3 x+5)^2 \cdot 3+(3 x+5)^3 \cdot 2(2 x-7) \cdot 2 \\
& =9(2 x-7)^2(3 x+5)^2+4(3 x+5)^3(2 x-7) \\
& =(2 x-7)(3 x+5)^2[9(2 x-7)+4(3 x+5)] \\
& =(2 x-7)(3 x+5)^2(18 x-63+12 x+20) \\
& =(2 x-7)(3 x+5)^2(30 x-43) \\
&
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
x^2 \sin x+\cos 2 x
\)
\(
\frac{d}{d x}\left(x^2 \sin x+\cos 2 x\right)=\frac{d}{d x}\left(x^2 \sin x\right)+\frac{d}{d x}(\cos 2 x)
\)
\(
\begin{aligned}
& =\left(x^2 \cos x+\sin x \cdot 2 x\right)+(-2 \sin 2 x) \\
& =x^2 \cos x+2 x \sin x-2 \sin 2 x
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\sin ^3 x \cos ^3 x
\)
\(
\frac{d}{d x}\left(\sin ^3 x \cos ^3 x\right)=\sin ^3 x \cdot \frac{d}{d x} \cos ^3 x+\cos ^3 x \cdot \frac{d}{d x}\left(\sin ^3 x\right)
\)
[Using Product Rule]
\(
\begin{aligned}
& =\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cdot \cos x \\
& =-3 \sin ^4 x \cos ^2 x+3 \cos ^4 x \sin ^2 x \\
& =3 \sin ^2 x \cos ^2 x\left(-\sin ^2 x+\cos ^2 x\right) \\
& =3 \sin ^2 x \cos ^2 x \cdot \cos 2 x \\
& =\frac{3}{4} \cdot 4 \sin ^2 x \cos ^2 x \cdot \cos 2 x=\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x \\
& =\frac{3}{4} \sin ^2 2 x \cdot \cos 2 x
\end{aligned}
\)
Differentiate each of the functions w. r. to \(x\)
\(
\frac{1}{a x^2+b x+c}
\)
\(
\begin{gathered}
\frac{d}{d x}\left(\frac{1}{a x^2+b x+c}\right)=\frac{\left(a x^2+b x+c\right) \frac{d}{d x}(1)-1 \cdot \frac{d}{d x}\left(a x^2+b x+c\right)}{\left(a x^2+b x+c\right)^2} \text { [Using quotient rule] } \\
\quad=\frac{\left(a x^2+b x+c\right) \times 0-(2 a x+b)}{\left(a x^2+b x+c\right)^2}=\frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2}
\end{gathered}
\)
Differentiate each of the functions with respect to ‘ \(x\) ‘
\(\cos \left(x^2+1\right)\)
Let \(f(x)=\cos \left(x^2+1\right) \dots(i)\)
\(
\Rightarrow \quad f(x+\Delta x)=\cos \left[(x+\Delta x)^2+1\right] \dots(ii)
\)
Subtracting eq. (i) from eq. (ii) we get
\(
f(x+\Delta x)-f(x)=\cos \left[(x+\Delta x)^2+1\right]-\cos \left(x^2+1\right)
\)
Dividing both sides by \(\Delta x\) we get
\(
\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{\cos \left[(x+\Delta x)^2+1\right]-\cos \left(x^2+1\right)}{\Delta x}
\)
\(
\begin{gathered}
\Rightarrow \operatorname{Lim}_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\cos \left[(x+\Delta x)^2+1\right]-\cos \left(x^2+1\right)}{\Delta x} \\
f^{\prime}(x)=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\cos \left[(x+\Delta x)^2+1\right]-\cos \left(x^2+1\right)}{\Delta x} \text { [By definitions of differentiations] } \\
{[\text { By definitions of differentiations] }} \\
-2 \sin \left[\frac{(x+\Delta x)^2+1+x^2+1}{2}\right]
\end{gathered}
\)
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\sin \left[\frac{(x+\Delta x)^2+1-x^2-1}{2}\right]}{\Delta x}
\)
\(
\left[\because \quad \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right]
\)
\(
-2 \sin \left[\frac{x^2+\Delta x^2+2 x \cdot \Delta x+x^2+2}{2}\right]
\)
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\sin \left[\frac{x^2+\Delta x^2+2 x \Delta x-x^2}{2}\right]}{\Delta x}
\)
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{-2 \sin \left[x^2+\frac{\Delta x^2}{2}+x \Delta x+1\right] \sin \left[\Delta x \frac{(\Delta x+2 x)}{2}\right]}{\Delta x}-2 \sin \left[x^2+\frac{\Delta x^2}{2}+x \Delta x+1\right]
\)
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\sin \left[\Delta x \frac{(\Delta x+2 x)}{2}\right]}{\Delta x\left[\frac{\Delta x+2 x}{2}\right]} \times\left(\frac{\Delta x+2 x}{2}\right)
\)
\(
\begin{gathered}
=\operatorname{Lim}_{\Delta x \rightarrow 0} \\
\therefore \Delta x\left[\frac{\Delta x+2 x}{2}\right] \rightarrow 0 \\
-2 \sin \left[x^2+\frac{\Delta x^2}{2}+x \Delta x+1\right] \cdot \frac{\sin \left[\Delta x \frac{(\Delta x+2 x)}{2}\right]}{\Delta x\left[\frac{\Delta x+2 x}{2}\right]}\times\left[\frac{\Delta x+2 x}{2}\right]
\end{gathered}
\)
Taking limit, we have
\(
=-2 \sin \left(x^2+1\right) \cdot 1 \cdot(x)=-2 x \sin \left(x^2+1\right) \quad\left[\because \quad \operatorname{Lim}_{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\)
Hence, the required answer is \(-2 x \sin \left(x^2+1\right)\).
Differentiate each of the functions with respect to ‘ \(x\) ‘
\(
\frac{a x+b}{a x+d}
\)
\(
\begin{aligned}
& \text { Let } \quad f(x)=\frac{a x+b}{c x+d} \dots(i) \\
& \Rightarrow \quad f(x+\Delta x)=\frac{a(x+\Delta x)+b}{c(x+\Delta x)+d} \dots(ii) \\
&
\end{aligned}
\)
Subtracting eq. (i) from eq. (ii) we get
\(
f(x+\Delta x)-f(x)=\frac{a(x+\Delta x)+b}{c(x+\Delta x)+d}-\frac{a x+b}{c x+d}
\)
Dividing both sides by \(\Delta x\) and take the limit, we get
\(
\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\frac{a(x+\Delta x)+b}{c(x+\Delta x)+d}-\frac{a x+b}{c x+d}}{\Delta x}
\)
\(
\begin{array}{r}
f^{\prime}(x)=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{(a x+a \Delta x+b)(c x+d)-(a x+b)(c x+c \Delta x+d)}{[c(x+\Delta x)+d](c x+d) \cdot \Delta x} \\
\text { [Using definition of differentiation] } \\
a c x^2+a c \Delta x \cdot x+b c x+a d x+a d \Delta x+b d
\end{array}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{-a c x^2-a c \Delta x \cdot x-a d x-b c x-b c \cdot \Delta x-b d}{(c x+c \Delta x+d)(c x+d) \cdot \Delta x} \\
& =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{(a d-b c) \Delta x}{(c x+c \Delta x+d)(c x+d) \cdot \Delta x} \\
& =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{(a d-b c)}{(c x+c \cdot \Delta x+d)(c x+d)}
\end{aligned}
\)
Taking limit, we have
\(
=\frac{(a d-b c)}{(c x+d)(c x+d)}=\frac{a d-b c}{(c x+d)^2}
\)
Differentiate each of the functions with respect to ‘ \(x\) ‘
\(
x^{\frac{2}{3}}
\)
Let
\(
\begin{aligned}
f(x) & =x^{2 / 3} \dots(i) \\
f(x+\Delta x) & =(x+\Delta x)^{2 / 3} \dots(ii)
\end{aligned}
\)
Subtracting eq. (i) from (ii) we get
\(
f(x+\Delta x)-f(x)=(x+\Delta x)^{2 / 3}-x^{2 / 3}
\)
Dividing both sides by \(\Delta x\) and take the limit.
\(
\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{(x+\Delta x)^{2 / 3}-x^{2 / 3}}{\Delta x}
\)
\(
\begin{aligned}
f^{\prime}(x) & =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x^{2 / 3}\left[1+\frac{\Delta x}{x}\right]^{2 / 3}-x^{2 / 3}}{\Delta x} \\
& =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x^{2 / 3}\left[\left(1+\frac{\Delta x}{x}\right)^{2 / 3}-1\right]}{\Delta x} \\
& =\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x^{2 / 3}\left[\left(1+\frac{2}{3} \cdot \frac{\Delta x}{x}+\cdots\right)-1\right]}{\Delta x}
\end{aligned}
\)
[By definition of differentiation]
[Expanding by Binomial theorem and rejecting the higher powers of \(\Delta x\) as \(\Delta x \rightarrow 0\) ]
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x^{2 / 3} \cdot \frac{2}{3} \cdot \frac{\Delta x}{x}}{\Delta x}=\frac{2}{3} x^{2 / 3-1}=\frac{2}{3} x^{-1 / 3}
\)
Differentiate each of the functions with respect to ‘ \(x\) ‘
\(x \cos x\)
Let
\(
\begin{aligned}
y & =x \cos x \dots(i) \\
y+\Delta y & =(x+\Delta x) \cos (x+\Delta x) \dots(ii)
\end{aligned}
\)
Subtracting eq. (i) from eq. (ii) we get
\(
y+\Delta y-y=(x+\Delta x) \cos (x+\Delta x)-x \cos x
\)
\(
\Delta y=x \cos (x+\Delta x)+\Delta x \cos (x+\Delta x)-x \cos x
\)
Dividing both sides by \(\Delta x\) and take the limits,
\(
\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x \cos (x+\Delta x)-x \cos x+\Delta x \cos (x+\Delta x)}{\Delta x}
\)
\(
\frac{d y}{d x}=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x[\cos (x+\Delta x)-\cos x]}{\Delta x}+\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\Delta x \cos (x+\Delta x)}{\Delta x}\)
\(\left[\text { By definition } \quad \operatorname{Lim}_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{d y}{d x}\right]\)
\(
=\operatorname{Lim}_{\Delta x \rightarrow 0} \frac{x\left[-2 \sin \frac{(x+\Delta x+x)}{2} \cdot \sin \frac{(x+\Delta x-x)}{2}\right]}{\Delta x}+\operatorname{Lim}_{\Delta x \rightarrow 0} \cos (x+\Delta x)
\)
\(
=\operatorname{Lim}_{\substack{\Delta x \rightarrow 0 \\ \therefore \frac{\Delta x}{2} \rightarrow 0}} \frac{x\left[-2 \sin \left(x+\frac{\Delta x}{2}\right) \cdot \sin \frac{\Delta x}{2}\right]}{2 \times \frac{\Delta x}{2}}+\operatorname{Lim}_{\Delta x \rightarrow 0} \cos (x+\Delta x)
\)
\(
\therefore \quad \frac{\Delta x}{2} \rightarrow 0 \quad \text { Taking the limits, we have }
\)
\(
\begin{aligned}
& =x[-\sin x]+\cos x \\
& =-x \sin x+\cos x
\end{aligned}
\)
\(
\left[\because \quad \underset{\frac{\Delta x}{2} \rightarrow 0}{\operatorname{Lim}} \frac{\sin \frac{\Delta x}{2}}{\frac{\Delta x}{2}}=1\right]
\)
Evaluate each of the following limits
\(\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}\)
\(
\begin{aligned}
& \operatorname{Lim}_{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y} \\
& =\operatorname{Lim}_{y \rightarrow 0} \frac{x \sec (x+y)+y \sec (x+y)-x \sec x}{y}
\end{aligned}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{y \rightarrow 0} \frac{[x \sec (x+y)-x \sec x]}{y}+\operatorname{Lim}_{y \rightarrow 0} \frac{y \sec (x+y)}{y} \\
& =\operatorname{Lim}_{y \rightarrow 0} \frac{x[\sec (x+y)-\sec x]}{y}+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y) \\
& =\operatorname{Lim}_{y \rightarrow 0} \frac{x\left[\frac{1}{\cos (x+y)}-\frac{1}{\cos x}\right]}{y}+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y)
\end{aligned}
\)
\(
=\operatorname{Lim}_{y \rightarrow 0} x\left[\frac{\cos x-\cos (x+y)}{y \cdot \cos (x+y) \cdot \cos x}\right]+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y)
\)
\(
=\operatorname{Lim}_{y \rightarrow 0} \frac{x\left[\begin{array}{l}
-2 \sin \left(\frac{x+x+y}{2}\right) \\
\cdot \sin \left(\frac{x-x-y}{2}\right)
\end{array}\right]}{y \cos (x+y) \cdot \cos x}+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y)
\)
\(
\begin{aligned}
& =\frac{x\left[-2 \sin \left(x+\frac{y}{2}\right) \cdot \sin \left(-\frac{y}{2}\right)\right]}{\cos (x+y) \cdot \cos x \cdot y}+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y) \\
& =\operatorname{Lim}_{\substack{y \rightarrow 0 \\
\therefore \frac{y}{2} \rightarrow 0}} x\left[\frac{\left[2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{y}{2}\right)\right]}{\cos (x+y) \cdot \cos x \cdot\left(\frac{y}{2}\right) \cdot 2}\right]+\operatorname{Lim}_{y \rightarrow 0} \sec (x+y)
\end{aligned}
\)
\(\therefore\) Taking the limits we have
\(
\begin{aligned}
& =x\left[\sin x \cdot \frac{1}{\cos x \cdot \cos x}\right]+\sec x \\
& =x \sec x \tan x+\sec x=\sec x(x \tan x+1)
\end{aligned}
\)
Evaluate each of the following limits
\(
\lim _{x \rightarrow 0} \frac{(\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha x)}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x
\)
\(
\begin{aligned}
& \text { Given, } \operatorname{Lim}_{x \rightarrow 0} \frac{[\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha \cdot x]}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{[2 \sin \alpha x \cdot \cos \beta x+\sin 2 \alpha \cdot x] \cdot x}{2 \sin (\alpha+\beta) x \cdot \sin (\alpha-\beta) x}
\end{aligned}
\)
\(
\left[\begin{array}{l}
\because \quad \sin C+\sin D=2 \sin \frac{C+D}{2} \cdot \cos \frac{C-D}{2} \\
\cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}
\end{array}\right]
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0} \frac{[2 \sin \alpha x \cdot \cos \beta x+2 \sin \alpha x \cdot \cos \alpha x] \cdot x}{2 \sin (\alpha+\beta) x \cdot \sin (\alpha-\beta) x} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin \alpha x(\cos \beta x+\cos \alpha x) \cdot x}{2 \sin (\alpha+\beta) x \cdot \sin (\alpha-\beta) x} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\sin \alpha x\left[2 \cos \left(\frac{\alpha+\beta}{2}\right) x \cdot \cos \left(\frac{\alpha-\beta}{2}\right) x\right] \cdot x}{\sin (\alpha+\beta) x \cdot \sin (\alpha-\beta) x} \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\sin \alpha x\left[2 \cos \left(\frac{\alpha+\beta}{2}\right) x \cdot \cos \left(\frac{\alpha-\beta}{2}\right) x\right] \cdot x}{2 \sin \left(\frac{\alpha+\beta}{2}\right) x \cdot \cos \left(\frac{\alpha+\beta}{2}\right) x} \\
& \cdot 2 \sin \left(\frac{\alpha-\beta}{2}\right) x \cdot \cos \left(\frac{\alpha-\beta}{2}\right) x
\end{aligned}
\)
\(
\left[\begin{array}{l}
\because \quad \cos C +\cos D =2 \cos \frac{ C + D }{2} \cdot \cos \frac{ C – D }{2} \\
\text { and } \sin 2 x=2 \sin x \cos x
\end{array}\right]
\)
\(
=\operatorname{Lim}_{x \rightarrow 0} \frac{\sin \alpha x \cdot x}{2 \sin \left(\frac{\alpha+\beta}{2}\right) x \sin \left(\frac{\alpha-\beta}{2}\right) \cdot x}
\)
\(
=\operatorname{Lim}_{x \rightarrow 0} \frac{1}{2} \frac{\frac{\sin \alpha x}{\alpha x} \cdot(\alpha x) \cdot x}{\left[\frac{\sin \left(\frac{\alpha+\beta}{2}\right) x}{\left(\frac{\alpha+\beta}{2}\right) \cdot x} \times\left(\frac{\alpha+\beta}{2}\right) \cdot x\right]\left[\frac{\sin \left(\frac{\alpha-\beta}{2}\right) \cdot x}{\left(\frac{\alpha-\beta}{2}\right) \cdot x} \times \frac{(\alpha-\beta)}{2} \cdot x\right]}
\)
\(
\begin{aligned}
& =\frac{1}{2} \cdot \frac{\alpha x^2}{\left(\frac{\alpha+\beta}{2}\right) x \cdot\left(\frac{\alpha-\beta}{2}\right) x} \\
& =\frac{1}{2}\left[\frac{\alpha}{\left(\frac{\alpha+\beta}{2}\right)\left(\frac{\alpha-\beta}{2}\right)}\right] \\
& =\frac{1}{2} \cdot \frac{4 \alpha}{\alpha^2-\beta^2}=\frac{2 \alpha}{\alpha^2-\beta^2}
\end{aligned}
\)
Evaluate each of the following limits
\(
\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}
\)
\(
\begin{aligned}
& \text { Given, } \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\tan ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} \\
& =\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\tan x\left(\tan ^2 x-1\right)}{\cos \left(x+\frac{\pi}{4}\right)}=\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \tan x \cdot \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}}\left[\frac{-\left(1-\tan ^2 x\right)}{\cos \left(x+\frac{\pi}{4}\right)}\right] \\
& =-1 \times \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{(1-\tan x)(1+\tan x)}{\cos \left(x+\frac{\pi}{4}\right)} \\
& =\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}}-(1+\tan x) \cdot \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}}\left(\frac{1-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}\right) \\
& =-(1+1) \cdot \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{(\cos x-\sin x)}{\cos x \cdot \cos \left(x+\frac{\pi}{4}\right)}
\end{aligned}
\)
\(
\begin{aligned}
& =-2 \times \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x\right)}{\cos x \cdot \cos \left(x+\frac{\pi}{4}\right)} \\
& =-2 \sqrt{2} \operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{\left[\cos \frac{\pi}{4} \cdot \cos x-\sin \frac{\pi}{4} \sin x\right]}{\cos x \cdot \cos \left(x+\frac{\pi}{4}\right)} \\
& =\operatorname{Lim}_{x \rightarrow \frac{\pi}{4}} \frac{-2 \sqrt{2} \cdot \cos \left(x+\frac{\pi}{4}\right)}{\cos x \cdot \cos \left(x+\frac{\pi}{4}\right)}=\frac{-2 \sqrt{2}}{\cos \frac{\pi}{4}} \quad \text { (Taking limit) } \\
& =\frac{-2 \sqrt{2}}{\frac{1}{\sqrt{2}}}=-2 \times 2=-4
\end{aligned}
\)
Evaluate each of the following limits
\(
\lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}
\)
Given, \(\operatorname{Lim}_{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow \pi} \frac{\cos ^2 \frac{x}{4}+\sin ^2 \frac{x}{4}-2 \sin \frac{x}{4} \cdot \cos \frac{x}{4}}{\left(\cos ^2 \frac{x}{4}-\sin ^2 \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)} \\
& {\left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]} \\
& =\operatorname{Lim}_{x \rightarrow \pi} \frac{\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)^2}{\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right) \cdot\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right) \cdot\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)} \\
& =\operatorname{Lim}_{x \rightarrow \pi} \frac{1}{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)} \\
&
\end{aligned}
\)
Taking limits we have
\(
=\frac{1}{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{1}{\frac{2}{\sqrt{2}}}=\frac{1}{\sqrt{2}}
\)
\(
\text { Does the } \lim _{x \rightarrow 4} \frac{|x-4|}{x-4} \text { exists? }
\)
\(
\begin{aligned}
& \text { Given } \operatorname{Lim}_{x \rightarrow 4} \frac{|x-4|}{x-4} \\
& LHL =\operatorname{Lim}_{x \rightarrow 4^{-}} \frac{-(x-4)}{x-4}=-1 \quad[\because \quad|x-4|=-(x-4) \text { if } x<4] \\
& RHL =\operatorname{Lim}_{x \rightarrow 4^{+}} \frac{x-4}{x-4}=1 \quad[\because|x-4|=(x-4) \text { if } x>4] \\
&
\end{aligned}
\)
Since \(LHL \neq RHL\)
Hence, the limit does not exist.
\(
\text { Let } f(x)=\left\{\begin{array}{cr}
\frac{k \cos x}{\pi-2 x} & \text { when } x \neq \frac{\pi}{2} \\
3 & x=\frac{\pi}{2}
\end{array}\right.
\)
\(
\text { and if } \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \text {, } \text { find the value of } k \text {. }
\)
Given,
\(
f(x)=\left\{\begin{aligned}
\frac{k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\
3, & x=\frac{\pi}{2}
\end{aligned}\right.
\)
\(
\begin{aligned}
\operatorname{LHL} f(x) & =\operatorname{Lim}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=\operatorname{Lim}_{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)} \\
& =\operatorname{Lim}_{h \rightarrow 0} \frac{k \sin h}{\pi-\pi+2 h}=\operatorname{Lim}_{h \rightarrow 0} \frac{k \sin h}{2 h} \\
& =\frac{k}{2} \cdot 1=\frac{k}{2} \left[\because \quad \operatorname{Lim}_{h \rightarrow 0} \frac{\sin h}{h}=1\right]
\end{aligned}
\)
\(
\operatorname{RHL} f(x)=\operatorname{Lim}_{x \rightarrow \frac{\pi^{+}}{2}} \frac{k \cos x}{\pi-2 x}=\operatorname{Lim}_{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{h \rightarrow 0} \frac{-k \sin h}{\pi-\pi-2 h} \\
& =\operatorname{Lim}_{h \rightarrow 0} \frac{-k \sin h}{-2 h}=\frac{k}{2} \quad\left[\because \quad \operatorname{Lim}_{h \rightarrow 0} \frac{\sin h}{h}=1\right]
\end{aligned}
\)
we are given that \(\operatorname{Lim}_{x \rightarrow \frac{\pi}{2}} f(x)=3\)
So, \(\frac{k}{2}=3 \Rightarrow k=6\)
\(
\text { Let } f(x)=\left\{\begin{array}{cc}
x+2 & x \leq-1 \\
c x^2 & x>-1
\end{array} \text {, find ‘ } c \text { ‘ if } \lim _{x \rightarrow-1} f(x)\right. \text { exists. }
\)
\(
\text { Given, } \quad f(x)=\left\{\begin{aligned}
x+2, & x \leq-1 \\
c x^2, & x>-1
\end{aligned}\right.
\)
\(
\begin{aligned}
LHL =\operatorname{Lim}_{x \rightarrow-1^{-}} f(x) & =\operatorname{Lim}_{x \rightarrow-1^{-}}(x+2) \\
& =\operatorname{Lim}_{h \rightarrow 0}(-1-h+2)=\operatorname{Lim}_{h \rightarrow 0}(1-h)=1
\end{aligned}
\)
\(
RHL =\operatorname{Lim}_{x \rightarrow-1^{+}} c x^2=\operatorname{Lim}_{h \rightarrow 0} c(-1+h)^2=c
\)
Since the limits exist.
\(
\begin{aligned}
& \therefore \quad \text { LHL }=\text { RHL } \\
& \therefore \quad c=1 \\
&
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2 x-3)}{x^3-3 x^2+2 x}\right]
\)
We have
\(
\begin{aligned}
\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2 x-3)}{x^3-3 x^2+2 x}\right] & =\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2 x-3)}{x(x-1)(x-2}\right] \\
& =\lim _{x \rightarrow 2}\left[\frac{x(x-1)-2(2 x-3)}{x(x-1)(x-2)}\right] \\
& =\lim _{x \rightarrow 2}\left[\frac{x^2-5 x+6}{x(x-1)(x-2)}\right] \\
& =\lim _{x \rightarrow 2}\left[\frac{(x-2)(x-3)}{x(x-1)(x-2)}\right][x-2 \neq 0] \\
& =\lim _{x \rightarrow 2}\left[\frac{x-3}{x(x-1)}\right]=\frac{-1}{2}
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}
\)
Put \(y=2+x\) so that when \(x \rightarrow 0, y \rightarrow 2\). Then
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x} & =\lim _{y \rightarrow 2} \frac{y^{\frac{1}{2}}-2^{\frac{1}{2}}}{y-2} \\
& =\frac{1}{2}(2)^{\frac{1}{2}-1}=\frac{1}{2} \cdot 2^{-\frac{1}{2}}=\frac{1}{2 \sqrt{2}}
\end{aligned}
\)
Find the positive integer \(n[latex] so that [latex]\lim _{x \rightarrow 3} \frac{x^n-3^n}{x-3}=108\).
We have
\(
\lim _{x \rightarrow 3} \frac{x^n-3^n}{x-3}=n(3)^{n-1}
\)
Therefore,
\(
n(3)^{n-1}=108=4(27)=4(3)^{4-1}
\)
Comparing, we get
\(
n=4
\)
\(
\text { Evaluate } \lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)
\)
Put \(y=\frac{\pi}{2}-x\). Then \(y \rightarrow 0[latex] as [latex]x \rightarrow \frac{\pi}{2}\). Therefore
\(
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x) & =\lim _{y \rightarrow 0}\left[\sec \left(\frac{\pi}{2}-y\right)-\tan \left(\frac{\pi}{2}-y\right)\right] \\
& =\lim _{y \rightarrow 0}(\operatorname{cosec} y-\cot y) \\
& =\lim _{y \rightarrow 0}\left(\frac{1}{\sin y}-\frac{\cos y}{\sin y}\right) \\
& =\lim _{y \rightarrow 0}\left(\frac{1-\cos y}{\sin y}\right)
\end{aligned}
\)
\(
\begin{aligned}
=\lim _{y \rightarrow 0} \frac{2 \sin ^2 \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}} & \binom{\text { since, } \sin ^2 \frac{y}{2}=\frac{1-\cos y}{2}}{\sin y=2 \sin \frac{y}{2} \cos \frac{y}{2}} \\
& =\lim _{\frac{y}{2} \rightarrow 0} \tan \frac{y}{2}=0
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}
\)
We have
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x} & =\lim _{x \rightarrow 0} \frac{2 \cos \frac{(2+x+2-x)}{2} \sin \frac{(2+x-2+x)}{2}}{x} \\
& =\lim _{x \rightarrow 0} \frac{2 \cos 2 \sin x}{x} \\
& =2 \cos 2 \lim _{x \rightarrow 0} \frac{\sin x}{x}=2 \cos 2\left(\text { as } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)
\end{aligned}
\)
Find the derivative of \(f(x)=a x+b\), where \(a\) and \(b\) are non-zero constants, by first principle.
By definition,
\(
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{a(x+h)+b-(a x+b)}{h}=\lim _{h \rightarrow 0} \frac{a h}{h}=a
\end{aligned}
\)
Find the derivative of \(f(x)=a x^2+b x+c\), where \(a, b\) and \(c\) are none-zero constant, by first principle.
By definition,
\(
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{a(x+h)^2+b(x+h)+c-a x^2-b x-c}{h} \\
=\lim _{h \rightarrow 0} \frac{b h+a h^2+2 a x h}{h} & =\lim _{h \rightarrow 0} a h+2 a x+b=b+2 a x
\end{aligned}
\)
Find the derivative of \(f(x)=x^3\), by first principle.
\(
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{(x+h)^3-x^3}{h} \\
& =\lim _{h \rightarrow 0} \frac{x^3+h^3+3 x h(x+h)-x^3}{h} \\
& =\lim _{h \rightarrow 0}\left(h^2+3 x(x+h)\right)=3 x^2
\end{aligned}
\)
Find the derivative of \(f(x)=\frac{1}{x}\) by first principle.
By definition
\(
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{x+h}-\frac{1}{x}\right) \\
& =\lim _{h \rightarrow 0} \frac{-h}{h(x+h) x}=\frac{-1}{x^2}
\end{aligned}
\)
Find the derivative of \(f(x)=\sin x\), by first principle.
By Definition
\(
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h} \\
& =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}}{2 \cdot \frac{h}{2}} \\
& =\lim _{h \rightarrow 0} \cos \frac{(2 x+h)}{2} \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \\
& =\cos x \cdot 1=\cos x
\end{aligned}
\)
Find the derivative of \(f(x)=x^n[latex], where [latex]n\) is positive integer, by first principle.
By definition,
\(
\begin{aligned}
f^{\prime}(x) & =\frac{f(x+h)-f(x)}{h} \\
& =\frac{(x+h)^n-x^n}{h}
\end{aligned}
\)
Using Binomial theorem, we have \((x+h)^n={ }^n C _0 x^n+{ }^n C _1 x^{n-1} h+\ldots+{ }^n C _n h^n\)
Thus,
\(
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{(x+h)^n-x^n}{h} \\
& =\lim _{h \rightarrow 0} \frac{h\left(n x^{n-1}+\ldots+h^{n-1}\right]}{h}=n x^{n-1} .
\end{aligned}
\)
Find the derivative of \(2 x^4+x\).
Let \(y=2 x^4+x\)
Differentiating both sides with respect to \(x\), we get
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(2 x^4\right)+\frac{d}{d x}(x) \\
& =2 \times 4 x^{4-1}+1 x^0
\end{aligned}
\)
\(
=8 x^3+1
\)
Therefore,
\(
\frac{d}{d x}\left(2 x^4+x\right)=8 x^3+1
\)
Find the derivative of \(x^2 \cos x\).
Let \(y=x^2 \cos x\)
Differentiating both sides with respect to \(x\), we get
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(x^2 \cos x\right) \\
& =x^2 \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}\left(x^2\right) \\
& =x^2(-\sin x)+\cos x(2 x) \\
& =2 x \cos x-x^2 \sin x
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}
\)
Note that
\(
\begin{array}{r}
2 \sin ^2 x+\sin x-1=(2 \sin x-1)(\sin x+1) \\
2 \sin ^2 x-3 \sin x+1=(2 \sin x-1)(\sin x-1)
\end{array}
\)
Therefore,
\(
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1} & =\lim _{x \rightarrow \frac{\pi}{6}} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)} \\
& \left.=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin x+1}{\sin x-1} \quad \text { (as } 2 \sin x-1 \neq 0\right) \\
& =\frac{1+\sin \frac{\pi}{6}}{\sin \frac{\pi}{6}-1}=-3
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\tan x-\sin x}{\sin ^3 x}
\)
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{\sin ^3 x} & =\lim _{x \rightarrow 0} \frac{\sin x\left(\frac{1}{\cos x}-1\right)}{\sin ^3 x} \\
& =\lim _{x \rightarrow 0} \frac{1-\cos x}{\cos x \sin ^2 x} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\cos x\left(4 \sin ^2 \frac{x}{2} \cdot \cos ^2 \frac{x}{2}\right)}=\frac{1}{2} .
\end{aligned}
\)
\(
\text { Evaluate } \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}
\)
\(
\text { We have } \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \times \frac{\sqrt{a+2 x}+\sqrt{3 x}}{\sqrt{a+2 x}+\sqrt{3 x}} \\
& =\lim _{x \rightarrow a} \frac{a+2 x-3 x}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})}
\end{aligned}
\)
\(
\begin{array}{r}
=\lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})} \\
=\lim _{x \rightarrow a} \frac{(a-x)[\sqrt{3 a+x}+2 \sqrt{x}]}{(\sqrt{a+2 x}+\sqrt{3 x})(3 a+x-4 x)}
\end{array}
\)
\(
=\frac{4 \sqrt{a}}{3 \times 2 \sqrt{3 a}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9} .
\)
\(
\text { Evaluate } \lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{\cos c x-1}
\)
We have \(\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{(a+b)}{2} x\right) \sin \frac{(a-b) x}{2}}{2 \frac{\sin ^2 c x}{2}}\)
\(
\begin{gathered}
=\lim _{x \rightarrow 0} \frac{2 \sin \frac{(a+b) x}{2} \cdot \sin \frac{(a-b) x}{2}}{x^2} \cdot \frac{x^2}{\sin ^2 \frac{c x}{2}} \\
=\lim _{x \rightarrow 0} \frac{\sin \frac{(a+b) x}{2}}{\frac{(a+b) x}{2} \cdot\left(\frac{2}{a+b}\right)} \cdot \frac{\sin \frac{(a-b) x}{2}}{\frac{(a-b) x}{2} \cdot \frac{2}{a-b}} \cdot \frac{\left(\frac{c x}{2}\right)^2 \times \frac{4}{c^2}}{\sin ^2 \frac{c x}{2}} \\
=\left(\frac{a+b}{2} \times \frac{a-b}{2} \times \frac{4}{c^2}\right)=\frac{a^2-b^2}{c^2}
\end{gathered}
\)
\(
\text { Evaluate } \lim _{h \rightarrow 0} \frac{(a+h)^2 \sin (a+h)-a^2 \sin a}{h}
\)
\(
\text { We have } \lim _{h \rightarrow 0} \frac{(a+h)^2 \sin (a+h)-a^2 \sin a}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\left(a^2+h^2+2 a h\right)[\sin a \cos h+\cos a \sin h]-a^2 \sin a}{h} \\
& =\lim _{h \rightarrow 0}\left[\frac{a^2 \sin a(\cos h-1)}{h}+\frac{a^2 \cos a \sin h}{h}+(h+2 a)(\sin a \cos h+\cos a \sin h)\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0}\left[\frac{a^2 \sin a\left(-2 \sin ^2 \frac{h}{2}\right)}{\frac{h^2}{2}} \cdot \frac{h}{2}\right]+\lim _{h \rightarrow 0} \frac{a^2 \cos a \sin h}{h}+\lim _{h \rightarrow 0}(h+2 a) \sin (a+h) \\
& =a^2 \sin a \times 0+a^2 \cos a(1)+2 a \sin a \\
& =a^2 \cos a+2 a \sin a .
\end{aligned}
\)
Find the derivative of \(f(x)=\tan (a x+b)\), by first principle.
\(
\text { We have } f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\tan (a(x+h)+b)-\tan (a x+b)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\frac{\sin (a x+a h+b)}{\cos (a x+a h+b)}-\frac{\sin (a x+b)}{\cos (a x+b)}}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sin (a x+a h+b) \cos (a x+b)-\sin (a x+b) \cos (a x+a h+b)}{h \cos (a x+b) \cos (a x+a h+b)} \\
& =\lim _{h \rightarrow 0} \frac{a \sin (a h)}{a \cdot h \cos (a x+b) \cos (a x+a h+b)}
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{a}{\cos (a x+b) \cos (a x+a h+b)} \lim _{a h \rightarrow 0} \frac{\sin a h}{a h}[\text { as } h \rightarrow 0 a h \rightarrow 0] \\
& =\frac{a}{\cos ^2(a x+b)}=a \sec ^2(a x+b) .
\end{aligned}
\)
Find the derivative of \(f(x)=\sqrt{\sin x}\), by first principle.
By definition,
\(
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sqrt{\sin (x+h)}-\sqrt{\sin x}}{h} \\
& =\lim _{h \rightarrow 0} \frac{(\sqrt{\sin (x+h)}-\sqrt{\sin x})(\sqrt{\sin (x+h)}+\sqrt{\sin x})}{h(\sqrt{\sin (x+h)}+\sqrt{\sin x})} \\
& =\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h(\sqrt{\sin (x+h)}+\sqrt{\sin x})} \\
& =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}}{2 \cdot \frac{h}{2}(\sqrt{\sin (x+h)}+\sqrt{\sin x})} \\
& =\frac{\cos x}{2 \sqrt{\sin x}}=\frac{1}{2} \cot x \sqrt{\sin x}
\end{aligned}
\)
\(
\text { Find the derivative of } \frac{\cos x}{1+\sin x} \text {. }
\)
Let \(y=\frac{\cos x}{1+\sin x}\)
Differentiating both sides with respects to \(x\), we get
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\cos x}{1+\sin x}\right) \\
& =\frac{(1+\sin x) \frac{d}{d x}(\cos x)-\cos x \frac{d}{d x}(1+\sin x)}{(1+\sin x)^2} \\
& =\frac{(1+\sin x)(-\sin x)-\cos x(\cos x)}{(1+\sin x)^2}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{-\sin x-\sin ^2 x-\cos ^2 x}{(1+\sin x)^2} \\
& =\frac{-(1+\sin x)}{(1+\sin x)^2}=\frac{-1}{1+\sin x}
\end{aligned}
\)
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