Quiz

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow-1}\left[1+x+x^2+\ldots+x^{10}\right]=1+(-1)+(-1)^2+\ldots+(-1)^{10} \\
&=1-1+1 \ldots+1=1 .
\end{aligned}
\)

Hint

\(
\text { We have } \lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100}=\frac{2}{101}
\)

Hint

Evaluating the function at 2, we get it of the form \(\frac{0}{0}\).
Hence \(\lim _{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x}=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^2}\)
\(
=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0}
\)
which is not defined.

Hint

Put \(y=1+x\), so that \(y \rightarrow 1\) as \(x \rightarrow 0\).
Then
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} & =\lim _{y \rightarrow 1} \frac{\sqrt{y}-1}{y-1} \\
& =\lim _{y \rightarrow 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1} \\
& =\frac{1}{2}(1)^{\frac{1}{2}-1} \text { (by the remark above) }=\frac{1}{2}
\end{aligned}
\)

Hint

\(\lim _{x \rightarrow 0} {\sin x}\approx{x}\)

Therefore \(
\lim _{x \rightarrow 0} \frac{\sin x}{x}\approx x/x=1
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{1-\cos x}{x}= & \lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{x}{2}\right)}{x}=\lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \sin \left(\frac{x}{2}\right) \\
= & \lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \lim _{x \rightarrow 0} \sin \left(\frac{x}{2}\right)=1.0=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x} & =\lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x} \cdot \frac{2 x}{\sin 2 x} \cdot 2\right] \\
& =2 \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div\left[\frac{\sin 2 x}{2 x}\right] \\
& =2 \cdot \lim _{4 x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div \lim _{2 x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right] \\
& =2.1 .1=2(\text { as } x \rightarrow 0,4 x \rightarrow 0 \text { and } 2 x \rightarrow 0)
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} \frac{\tan x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x \cos x}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{\cos x}=1.1=1
\)

Hint

\(
\begin{aligned}
f^{\prime}(-1) & =\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left[2(-1+h)^2+3(-1+h)-5\right]-\left[2(-1)^2+3(-1)-5\right]}{h} \\
& =\lim _{h \rightarrow 0} \frac{2 h^2-h}{h}=\lim _{h \rightarrow 0}(2 h-1)=2(0)-1=-1
\end{aligned}
\)

Hint

A direct application of the above Theorem 6 tells that the derivative of the above function is \(1+2 x+3 x^2+\ldots+50 x^{49}\). At \(x=1\) the value of this function equals
\(
1+2(1)+3(1)^2+\ldots+50(1)^{49}=1+2+3+\ldots+50=\frac{(50)(51)}{2}=1275
\)

Hint

\(
\begin{aligned}
\frac{d f(x)}{d x} & =\frac{d}{d x}(\sin x \sin x) \\
& =(\sin x)^{\prime} \sin x+\sin x(\sin x)^{\prime} \\
& =(\cos x) \sin x+\sin x(\cos x) \\
& =2 \sin x \cos x=\sin 2 x
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} \frac{\sin (2+x) \sin (2-x)}{x}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{2 \cos \frac{(2+x+2-x)}{2} \sin \frac{2+x-2+x}{2}}{x} \\
& =\lim _{x \rightarrow 0} \frac{2 \cos 2 \sin x}{x} \\
& =2 \cos 2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \\
& =2 \cos 2(1) \\
& \text { (As, } \left.\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)
\end{aligned}
\)
Hence, \(\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=2 \cos 2\)

Hint

\(
\lim _{x \rightarrow 0} \frac{\cos (2 x)-1}{\cos x-1}=\frac{0}{0}
\)
(Recall the trigonometry identity \(\cos (2 x)=2 \cos ^2 x-1\) )
\(
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(2 \cos ^2 x-1\right)-1}{\cos x-1} \\
& =\lim _{x \rightarrow 0} \frac{2 \cos ^2 x-2}{\cos x-1} \\
& =\lim _{x \rightarrow 0} \frac{2\left(\cos ^2 x-1\right)}{\cos x-1}
\end{aligned}
\)
(The numerator is the difference of squares. Factor it.)
\(
=\lim _{x \rightarrow 0} \frac{2(\cos x-1)(\cos x+1)}{\cos x-1}
\)
(Divide out the factors \(\cos x-1\), the factors which are causing the indeterminate form \(\frac{0}{0}\). Now the limit can be computed.)
\(
\begin{gathered}
=\lim _{x \rightarrow 0} 2(\cos x+1) \\
=2(\cos 0+1) \\
=4 .
\end{gathered}
\)

Hint

\(
\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}=\frac{0}{0}
\)
(Make the replacement \(h=x-\frac{\pi}{2}\) so that \(x=h+\frac{\pi}{2}\). Note that as \(x\) approaches \(\frac{\pi}{2}, h\) approaches 0 .)
\(
=\lim _{h \rightarrow 0} \frac{\tan (2 h+\pi)}{h}
\)
(Recall the well-known, but seldom-used, trigonometry identity \(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\).)
\(
\begin{gathered}
=\lim _{h \rightarrow 0} \frac{1}{h} \frac{\tan 2 h+\tan \pi}{1-\tan 2 h \tan \pi} \\
=\lim _{h \rightarrow 0} \frac{1}{h} \frac{\tan 2 h+\{0\}}{1-\tan 2 h\{0\}} \\
=\lim _{h \rightarrow 0} \frac{1}{h} \tan 2 h \\
=\lim _{h \rightarrow 0} \frac{1}{h} \frac{\sin 2 h}{\cos 2 h}
\end{gathered}
\)
(Recall the well-known trigonometry identity \(\sin 2 h=2 \sin h \cos h\). )
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{1}{h} \frac{2 \sin h \cos h}{\cos 2 h} \\
& =\lim _{h \rightarrow 0} \frac{\sin h}{h} \frac{2 \cos h}{\cos 2 h}
\end{aligned}
\)
(Recall that \(\lim _{h \rightarrow 0} \frac{\sin h}{h}=1\).)
\(
\begin{aligned}
& =\{1\} \frac{2 \cos 0}{\cos 0} \\
& =\{1\} \frac{2\{1\}}{\{1\}}=2
\end{aligned}
\)

Hint

Any number divided by infinity gives us 0.
Here, since the number 2 is divided by \(y\), as \(y\) approaches infinity, we get 0.

Hint

Since it is of the form \(\frac{\infty}{\infty}\), we use L’Hospital’s rule and differentiate the numerator and denominator
\(
\mathrm{L}=\lim _{x \rightarrow \infty} \frac{x^2-9}{x^2-3 x+2}
\)
On differentiating once, we get \(\lim _{x \rightarrow \infty} \frac{2 x}{2 x}\)
Which is equal to, \(\lim _{x \rightarrow \infty} 1=1\).

Hint

Explanation: \(\lim _{y \rightarrow 4} f(y)=y^2+6 y\)
\(
\begin{aligned}
& f(4)=4^2+6(4) \\
& f(4)=16+24 \\
& f(4)=40
\end{aligned}
\)

Hint

Divide both numerator and denominator by \(x^2\) and they apply the limit. You will get ratio of 1/1=1.

Hint

\(
\begin{aligned}
\lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0^{+}}(3 x-7) \\
& =0-7 \\
& =-7
\end{aligned}
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow 0^{-}} f(x) & =\lim _{x \rightarrow 0^{-}}(x+2) \\
& =0+2 \\
& =2
\end{aligned}
\)

Hint

\(\lim _{x \rightarrow 0} f(x)\) does not exist
since
\(
\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) .
\)
Recall from the previous two questions that
\(
\lim _{x \rightarrow 0^{-}} f(x)=2
\)
and
\(
\lim _{x \rightarrow 0^{+}} f(x)=-7 .
\)

Hint

\(
\lim _{x \rightarrow a} 17=17
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow-7} \sqrt{x+4} & =\sqrt{-7+4} \\
& =\sqrt{-3} \\
& =i \sqrt{3},
\end{aligned}
\)
where \(i=\sqrt{-1}\) is called the “imaginary number.”

Hint

\(
\begin{aligned}
\lim _{n \rightarrow \infty} a_n & =\lim _{n \rightarrow \infty}\left(2+\frac{1}{n}\right) \\
& =\lim _{n \rightarrow \infty} 2+\lim _{n \rightarrow \infty} \frac{1}{n} \\
& =2+0 \\
& =2 .
\end{aligned}
\)

Hint

First, divide both the numerator and denominator by \(x^3\) and let \(h=\frac{-1}{x}\). This allows us to rewrite the limit as
\(
\begin{aligned}
\lim _{x \rightarrow-\infty} \frac{5 x^3+4 x+7}{25-2 x^3} & =\lim _{h \rightarrow 0^{+}} \frac{5+4 h^2-7 h^3}{-25 h^3-2} \\
& =\frac{-5}{2} .
\end{aligned}
\)

Hint

Since
\(
\frac{\lim _{x \rightarrow \infty} e^x}{\lim _{x \rightarrow \infty}\left(x^2+4\right)}=\frac{\infty}{\infty},
\)
we apply l’Hôpital’s rule, which gives us
\(
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{e^x}{x^2+4} & =\lim _{x \rightarrow \infty} \frac{\left(e^x\right)^{\prime}}{\left(x^2+4\right)^{\prime}} \\
& =\lim _{x \rightarrow \infty} \frac{e^x}{2 x} .
\end{aligned}
\)
Since
\(
\frac{\lim _{x \rightarrow \infty} e^x}{\lim _{x \rightarrow \infty} 2 x}=\frac{\infty}{\infty},
\)
we apply l’Hôpital’s rule a second time:
\(
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{e^x}{2 x} & =\lim _{x \rightarrow \infty} \frac{\left(e^x\right)^{\prime}}{(2 x)^{\prime}} \\
& =\lim _{x \rightarrow \infty} \frac{e^x}{2} \\
& =\frac{\infty}{2} \\
& =\infty
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow-\infty} \frac{3 x-2}{e^{x^2}}\left(\frac{-\infty}{\infty}\right) = \lim _{x \rightarrow-\infty} \frac{3}{e^{x^2(2 x)}} \quad\left(\frac{3}{\text { large neg. }}\right) \\
& =0 .
\end{aligned}
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^x-1-x-x^2 / 2}{x^3}\left(\frac{0}{0}\right) {=} \lim _{x \rightarrow 0} \frac{e^x-1-x}{3 x^2}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow 0} \frac{e^x-1}{6 x}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow 0} \frac{e^x}{6}=\frac{e^0}{6}=\frac{1}{6} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
\lim _{x \rightarrow \infty} x e^{-x} & =\lim _{x \rightarrow \infty} \frac{x}{e^x}\left(\frac{\infty}{\infty}\right) \\
{=} \lim _{x \rightarrow \infty} \frac{1}{e^x} \quad\left(\frac{1}{\text { large pos. }}\right) \\
& =0 .
\end{aligned}
\)

Hint

First
\(
\begin{aligned}
\lim _{x \rightarrow 0^{+}} \cot 2 x & =\lim _{x \rightarrow 0^{+}} \frac{\cos 2 x}{\sin 2 x} \quad\left(\frac{\text { about } 1}{\text { small pos. }}\right) \\
& =\infty .
\end{aligned}
\)
Therefore, the given limit is indeterminate of type \(\infty \cdot 0\). We rewrite as a fraction and then use l’Hôpital’s rule:
\(
\begin{aligned}
\lim _{x \rightarrow 0^{+}}(\cot 2 x)(\sin 6 x) & =\lim _{x \rightarrow 0^{+}} \frac{\sin 6 x}{\tan 2 x}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow 0^{+}} \frac{6 \cos 6 x}{2 \sec ^2 2 x} \\
& =\frac{6}{2}=3 .
\end{aligned}
\)

Hint

As observed above, this limit is indeterminate of type \(\infty-\infty\). We combine the terms and then use l’Hôpital’s rule:
\(
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}^{-}}(\tan x-\sec x) & =\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\sin x}{\cos x}-\frac{1}{\cos x}\right) \\
& =\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\sin x-1}{\cos x}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\cos x}{-\sin x} \\
& =\frac{0}{-1}=0 .
\end{aligned}
\)

Hint

\(
\text { The limit is indeterminate of type } \infty-\infty \text {. We combine the terms and then use }
\)
l’Hoppital’s rule:
\(
\begin{aligned}
\lim _{x \rightarrow 1^{+}}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) & =\lim _{x \rightarrow 1^{+}} \frac{x \ln x-x+1}{(x-1) \ln x}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow 1^{+}} \frac{\ln x+x(1 / x)-1}{\ln x+(x-1)(1 / x)} \\
& =\lim _{x \rightarrow 1^{+}} \frac{\ln x}{\ln x+1-1 / x}\left(\frac{0}{0}\right) \\
& =\lim _{x \rightarrow 1^{+}} \frac{1 / x}{1 / x+1 / x^2} \\
& =\frac{1}{2} .
\end{aligned}
\)

Hint

As was noted above, this limit is of indeterminate type \(\infty^{\circ}\). We first find the limit of the natural logarithm of the given expression:
\(
\begin{aligned}
\lim _{x \rightarrow \infty} \ln x^{1 / x} & =\lim _{x \rightarrow \infty}(1 / x) \ln x \quad(0 \cdot \infty) \\
& =\lim _{x \rightarrow \infty} \frac{\ln x}{x} \quad\left(\frac{\infty}{\infty}\right) \\
{=} \lim _{x \rightarrow \infty} \frac{1 / x}{1} \\
& =0 .
\end{aligned}
\)
Therefore,
\(
\lim _{x \rightarrow \infty} x^{1 / x}=\lim _{x \rightarrow \infty} e^{\ln x^{1 / x}}=e^0=1,
\)
where we have used the inverse property of logarithms \(y=e^{\ln y}\) and then the previous computation. 

Hint

As was noted above, this limit is indeterminate of type \(0^0\). We first find the limit of the natural logarithm of the given expression:
\(
\begin{aligned}
\lim _{x \rightarrow 0^{+}} \ln (\tan x)^x & =\lim _{x \rightarrow 0^{+}} x \ln \tan x \\
& =\lim _{x \rightarrow 0^{+}} \frac{\ln \tan x}{x^{-1}}\left(\frac{-\infty}{\infty}\right) \\
{=} \lim _{x \rightarrow 0^{+}} \frac{\cot x \sec ^2 x}{-x^{-2}} \\
& =\lim _{x \rightarrow 0^{+}} \frac{-x^2}{\sin x \cos x}\left(\frac{0}{0}\right) \\
{=} \lim _{x \rightarrow 0^{+}} \frac{-2 x}{\cos ^2 x-\sin ^2 x} \\
& =\frac{0}{1}=0 .
\end{aligned}
\)
Therefore,
\(
\lim _{x \rightarrow 0^{+}}(\tan x)^x=\lim _{x \rightarrow 0^{+}} e^{\ln (\tan x)^x}=e^0=1 .
\)

Hint

As was noted above, this limit is indeterminate of type \(1^{\infty}\). We first find the limit of the natural logarithm of the given expression:
\(
\begin{aligned}
\lim _{x \rightarrow \infty} \ln \left(1+\frac{1}{x}\right)^x & =\lim _{x \rightarrow \infty} x \ln \left(1+\frac{1}{x}\right) \\
& =\lim _{x \rightarrow \infty} \frac{\ln (1+1 / x)}{x^{-1}}\left(\frac{0}{0}\right) \\
& {=} \lim _{x \rightarrow \infty} \frac{\frac{1}{1+1 / x}\left(-x^{-2}\right)}{-x^{-2}} \\
& =\lim _{x \rightarrow \infty} \frac{1}{1+1 / x}=1 .
\end{aligned}
\)
Therefore,
\(
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=\lim _{x \rightarrow \infty} e^{\ln \left(1+\frac{1}{x}\right)^x}=e^1=e .
\)

Hint

We have
\(
\begin{aligned}
\sqrt{x^2-3 x+2} \text { is real } & \Leftrightarrow x^2-3 x+2 \geq 0 \\
& \Leftrightarrow(x-1)(x-2) \geq 0 \\
& \Leftrightarrow x \leq 1 \quad \text { or } \quad x \geq 2 \dots(i)
\end{aligned}
\)
Again
\(
\begin{aligned}
\frac{1}{\sqrt{3+2 x-x^2}} \text { is defined } & \Leftrightarrow 3+2 x-x^2>0 \\
& \Leftrightarrow-(x+1)(x-3)>0 \\
& \Leftrightarrow(x+1)(x-3)<0 \\
& \Leftrightarrow-1<x<3 \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), it follows that \(f(x)\) is defined for \(x \in(-1,1] \cup[2,3)\).

Hint

The given graph is represented by the function
\(
f(x)= \begin{cases}\frac{1}{x+1}+1 & \text { for } x<-1 \\ x & \text { for } 1<x<2\end{cases}
\)

Hint

Since \(f\) is even, we have \(f(1)=f(-1)\). Therefore \(2+\lambda=2-\lambda\), which implies that \(\lambda=0\).

Hint

Since \(C_2\) is the reflection of \(C_1\) in the origin, we have
\(
\begin{aligned}
(x, y) \in C_1 & \Leftrightarrow(-x,-y) \in C_2 \\
& \Leftrightarrow(-x)^2-13(-x)+4(-y)=1 \\
& \Leftrightarrow x^2+13 x-4 y=1
\end{aligned}
\)

Hint

Clearly
\(
g(x)=\frac{x^2-9}{x+3}=x-3=f(x)
\)
for \(x \neq-3\) and \(f(-3)=-6\). Hence \(k=-6\).

Hint

We have
\(
\begin{aligned}
|x-1| \geq 1 & \Leftrightarrow x-1 \leq-1 \quad \text { or } \quad x-1 \geq 1 \\
& \Leftrightarrow x \leq 0 \quad \text { or } \quad x \geq 2
\end{aligned}
\)

Hint

We have that
\(
\begin{aligned}
f(x) \text { is defined } & \Leftrightarrow|[|x|-1]|-5>0 \\
& \Leftrightarrow[|x|-1]<-5 \text { or } \quad[|x|-1]>5 \\
& \Leftrightarrow|x|-1<-5 \text { or }|x|-1 \geq 6
\end{aligned}
\)
\(
\begin{aligned}
& \Leftrightarrow|x|<-4 \quad \text { or } \quad|x| \geq 7 \\
& \Leftrightarrow x \leq-7 \quad \text { or } \quad x \geq 7 \\
& \Leftrightarrow x \in(-\infty,-7] \cup[7, \infty)
\end{aligned}
\)

Hint

\(\operatorname{Cos}^{-1}[(|x|-3) / 2]\) is defined for
\(
\begin{aligned}
& -1 \leq \frac{|x|-3}{2} \leq 1 \\
\Rightarrow & -2 \leq|x|-3 \leq 2 \\
\Rightarrow & 1 \leq|x| \leq 5
\end{aligned}
\)
Therefore
\(
x \in[-5,-1] \cup[1,5] \dots(i)
\)
Again \(1 / \log _{10}(4-x)\) is defined for \(x<4\) and \(x \neq 3\). That is
\(
x \in(-\infty, 3) \cup(3,4) \dots(ii)
\)
Therefore, from Eqs. (i) and (ii), the domain of the given function is
\(
[-5,-1] \cup[1,3) \cup(3,4)
\)

Hint

We have
\(
\operatorname{Sin}^{-1} x \text { is defined for }-1 \leq x \leq 1 \dots(i)
\)
Now \(1 /[x]\) is defined when \([x] \neq 0\), that is
\(
x \notin[0,1) \dots(ii)
\)
Also \(1 / \sqrt{x+1}\) is defined only when \(x+1>0\), that is
\(
x>-1 \dots(iii)
\)
From Eqs. (i) – (ii), the domain of the given function is \((-1,0) \cup\{1\}\).

Hint

1. \(e^x\) is defined for all \(x\) (real or complex). In this context \(x\) is real.
2. \(\sin ^{-1}[(x / 2)-1]\) is defined for \(-1 \leq(x / 2)-1 \leq 1\), that is, \(0 \leq x \leq 4\).
3. \(x \geq[x] \Rightarrow \sqrt{x-[x]}\) is defined for all real \(x\). Therefore, \(\log (\sqrt{x-[x]})\) is defined when \(x\) is not an integer, because \(x=[x]\) when \(x\) is an integer.

Therefore from the above, the domain of the given function is \((0,4)-\{1,2,3\}\).

Hint

\(\log _{10}\left[\left(3 x-x^2\right) / 2\right]\) is defined only when
\(
\begin{aligned}
& \frac{3 x-x^2}{2}>0 \\
\Rightarrow & x^2-3 x<0 \\
\Rightarrow & 0<x<3 \dots(i)
\end{aligned}
\)
Now \(\sqrt{\log _{10}\left[\left(3 x-x^2\right) / 2\right]}\) is defined when
\(
\begin{aligned}
& \frac{3 x-x^2}{2} \geq 1 \\
\Rightarrow & x^2-3 x+2 \leq 0 \\
\Rightarrow & (x-1)(x-2) \leq 0 \\
\Rightarrow & 1 \leq x \leq 2 \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), the domain of \(f(x)\) is \([1,2]\).

Hint

Clearly \(f(0)=f(0+0)=f(0)+f(0)\) implies \(f(0)=0\). Also
\(
0=f(0)=f(x-x)=f(x+(-x))=f(x)+f(-x)
\)
Therefore
\(
f(-x)=-f(x) \dots(i)
\)
Case I: Suppose \(x\) is a positive integer. Now
\(
\begin{gathered}
f(2)=f(1+1)=f(1)+f(1)=2 f(1) \\
f(3)=f(2+1)=f(2)+f(1)=2 f(1)+f(1)=3 f(1)
\end{gathered}
\)
Therefore, by induction,
\(
f(x)=x f(1) \dots(ii)
\)
Case II: Suppose \(x\) is a negative integer, say \(x=-y\) where \(y\) is a positive integer. Now from Eqs. (i) and (ii) we have
\(
f(x)=f(-y)=-f(y)=-y f(1)=x f(1)
\)
Case III: Suppose \(m\) and \(n\) are integers and \(x=m / n\) and \(n\) is positive. Now
\(
\begin{aligned}
m f(1) & =f(m)[\text { By cases }(1) \text { and }(2)] \\
& =f(n x) \\
& =f(x+x+\cdots+n \text { times }) \\
& =f(x)+f(x)+\cdots \text { upto } n \text { times } \\
& =n f(x)
\end{aligned}
\)
Therefore
\(
f(x)=\left(\frac{m}{n}\right) f(1)
\)
Hence, \(f(x)=x f(1)\) for all rational numbers \(x\).

Note: In the above problem, if \(f\) is also continuous, then \(f(x)=x f(1)\) for all real \(x\) which we discuss later.

Hint

\(f(x)\) is defined when \(1+[x] \neq 0\).
\(
\begin{aligned}
1+[x]=0 & \Leftrightarrow[x]=-1 \\
& \Leftrightarrow-1 \leq x<0 \\
& \Leftrightarrow x \in[-1,0)
\end{aligned}
\)

Hint

The given function can be written as
\(
f(x)=\frac{x|x-3|}{(x-3)(x+2)|x|}
\)
Therefore, \(f\) is defined for \(x \neq-2,0,3\).

Hint

\(1-x \geq 0\) and \(x(1-x) \geq 0\) when \(0 \leq x \leq 1\). In this case, \(0 \leq 2 \sqrt{(1-x) x} \leq 1\).

Hint

3 does not belong to the range of \(f\) implies 2 also cannot belong to range of \(f\) because, if \(f(x)=2\) for some \(x \in R\). Then
\(
f(x+p)=\frac{2-5}{2-3}=3
\)
which is not in the range of \(f\). Hence 2 and 3 are not in the range of \(f\). If \(f(x+2 p)=f(x)\), this implies
\(
\begin{aligned}
f(x) & =f(x+p+p) \\
& =\frac{f(x+p)-5}{f(x+p)-3} \\
& =\frac{\frac{f(x)-5}{f(x)-3}-5}{\frac{f(x)-5}{f(x)-3}-3}
\end{aligned}
\)
\(
=\frac{-4 f(x)+10}{-2 f(x)+4}=\frac{2 f(x)-5}{f(x)-2}
\)
so that \([f(x)-2]^2=-1\) which is absurd. Therefore, \(2 p\) is not a period. Again
\(
\begin{aligned}
f(x+3 p) & =\frac{2 f(x+p)-5}{f(x+p)-2} \\
& =\frac{3 f(x)-5}{f(x)-1} \neq f(x)
\end{aligned}
\)
Now
\(
\begin{aligned}
f(x+4 p) & =f(x+3 p+p) \\
& =\frac{f(x+3 p)-5}{f(x+3 p)-3} \\
& =\frac{\frac{3 f(x)-5}{f(x)-1}-5}{\frac{3 f(x)-5}{f(x)-1}-3} \\
& =\frac{-2 f(x)}{-2}=f(x)
\end{aligned}
\)
Therefore \(4 p\) is a period.

Hint

Given that
\(
f(x+y)+f(x-y)=2 f(x) f(y) \forall x, y \in R
\)
Substituting \(x=0\) and \(y=0\), we have
\(
f(0)=(f(0))^2 \Rightarrow f(0)=0 \quad \text { or } 1
\)
If \(f(0)=0\), then
\(
f(x+0)+f(x-0)=2 f(x) f(0)
\)
and hence \(f(x)=0\) for all \(x \in R\) which contradicts the fact that \(f(\alpha)=-1\). Therefore
\(
f(0)=1
\)
Now, replacing \(x\) with \(x+2 \alpha\) and \(y\) with \(x-2 \alpha\) in the given relation, we have
\(
f(2 x)+f(4 \alpha)=2 f(x+2 \alpha) f(x-2 \alpha) \dots(1)
\)
Also in the given relation, if we put \(y=x\), then we have
\(
f(2 x)+f(0)=2 f(x) f(x)
\)
Therefore
\(
f(2 x)=2[f(x)]^2-1 \dots(2)
\)
In Eq. (2), if we replace \(x\) with \(2 \alpha\), then
\(
f(4 \alpha)=2[f(2 \alpha)]^2-1 \dots(3)
\)
But
\(
\begin{aligned}
x=y=\alpha & \Rightarrow f(2 \alpha)+f(0)=2[f(\alpha)]^2 \\
& \Rightarrow f(2 \alpha)=2[f(\alpha)]^2-1 \\
& =2(1)-1 \quad[\because f(\alpha)=-1] \\
& =1 \dots(4)
\end{aligned}
\)
From Eqs. (3) and (4) we have
\(
f(4 \alpha)=1 \dots(5)
\)
From Eqs. (1), (2) and (5), we get that
\(
f(x+2 \alpha) f(x-2 \alpha)=(f(x))^2 \dots(6)
\)
Similarly if we put \(y=2 \alpha\) in the given relation, we have
\(
f(x+2 \alpha)+f(x-2 \alpha)=2 f(x) \dots(7)
\)
From Eqs. (6) and (7), we have
\(
f(x-2 \alpha)=f(x+2 \alpha)=f(x)
\)
Therefore \(2 \alpha\) is a period of \(f(x)\).

Hint

Observe that \(f(x) \geq 1 / 2\) for all \(x \in R\). Now
\(
\begin{aligned}
& =\frac{1}{2}+\sqrt{\frac{1}{4}-f(x)+[f(x)]^2} \\
&
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2}+\left|\frac{1}{2}-f(x)\right| \\
& =\frac{1}{2}-\frac{1}{2}+f(x) \quad\left(\because f(x) \geq \frac{1}{2} \forall x\right) \\
& =f(x)
\end{aligned}
\)
Therefore, 2 is period of \(f(x)\).

Hint

\(\quad f(x+\pi)=\operatorname{Tan}^{-1}(\tan (\pi+x))=\operatorname{Tan}^{-1}(\tan x)=f(x)\)
This implies \(\pi\) is the least period of \(f(x)\).

Hint

Period of \(a \cos (a x+b)\) is \(2 \pi\) when \(a\) is an integer and the period of \(\cos (k x)=2 \pi / k\) when \(k>0\) integer. Therefore, period of \(f(x)\) is \(6 \pi\).

Hint

It can be seen that
\(
f(x)= \begin{cases}5 & \text { if } x \leq-1 \\ 3-2 x & \text { if }-1 \leq x \leq 2 \\ 2 x-5 & \text { if } x \geq 2\end{cases}
\)

Hint

The domain of \(f(x)=\sqrt{|x-1|-1}\) is \((-\infty, 0] \cup\) \([2, \infty)\) because \(|x-1| \geq 1\). Now

\(
x \leq 0 \Rightarrow f(x)=\sqrt{|x-1|-1}=\sqrt{-x}
\)
So the graph must be upper half the parabola \(y^2=-x\) with vertex at origin. Again
\(
x \geq 2 \Rightarrow y=\sqrt{x-2} \Rightarrow y^2=x-2
\)
which represents parabola in the upper half of the \(x\)-axis with vertex at \((2,0)\).

Hint

As per the graph, \(\lim _{x \rightarrow 3+0} f(x)=2\)

Hint

We have
\(
\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\left[3(x+h)^2-(x+h)\right]-\left(3 x^2-x\right)}{h}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{6 x h+3 h^2-h}{h} \\
& =\lim _{h \rightarrow 0}[6 x+3 h-1]=6 x-1
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 1-0} f(x) & =\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)=1 \\
\lim _{x \rightarrow 1+0} f(x) & =\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[2(1+h)-1] \\
& =\lim _{h \rightarrow 0}(2+2 h-1)=1
\end{aligned}
\)
Therefore
\(
\begin{aligned}
& \lim _{x \rightarrow 1-0} f(x)=\lim _{x \rightarrow 1+0} f(x)=1 \\
\Rightarrow & \lim _{x \rightarrow 1} f(x)=1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow 1-0} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[(1-h)-1]=0 \\
& \lim _{x \rightarrow 1+0} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[2(1+h)-1]=1 \\
& \lim _{x \rightarrow 1-0} f(x) \neq \lim _{x \rightarrow 1+0} f(x) \Rightarrow \lim _{x \rightarrow 1} f(x) \text { does not exist }
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
& \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _{h \rightarrow 0} \frac{(a+h)[a+h]-a[a]}{h} \\
& =\lim _{h \rightarrow 0} \frac{(a+h)[a]-a[a]}{h} \quad(\because a \text { is not an integer, } \\
& {[a+h]=[a] \text { for small values of } h \text { ) }} \\
& =\lim _{h \rightarrow 0}[a]=[a] \\
&
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} x\left(\frac{\sin x^2}{x^2}\right) \\
& =0 \times 1 \quad \text { (By Theorem 1.27) } \\
& =0
\end{aligned}
\)

Hint

\(
\begin{aligned}
f(x) & =\frac{(1+x)^{1 / 2}-1}{(1+x)^{1 / 3}-1} \\
& =\left(\frac{(1+x)^{1 / 2}-1}{1+x-1}\right) \div \frac{(1+x)^{1 / 3}-1}{1+x-1} \\
& =\frac{y^{1 / 2}-1}{y-1} \div \frac{y^{1 / 3}-1}{y-1}
\end{aligned}
\)
where \(y=1+x\). Now \(y \rightarrow 1\) as \(x \rightarrow 0\). Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\lim _{y \rightarrow 1}\left[\left(\frac{y^{1 / 2}-1}{y-1}\right) \div\left(\frac{y^{1 / 3}-1}{y-1}\right)\right] \\
& =\lim _{y \rightarrow 1}\left(\frac{y^{1 / 2}-1}{y-1}\right) \div \lim _{y \rightarrow 1}\left(\frac{y^{1 / 3}-1}{y-1}\right) \\
& =\frac{1}{2} \div \frac{1}{3} \quad \text { (By Theorem 1.26) } \\
& =\frac{3}{2}
\end{aligned}
\)

Hint

Let
\(
\begin{aligned}
f(x) & =\frac{\sin (a+x)-\sin (a-x)}{x} \\
& =\frac{2 \cos a \sin x}{x} \\
& =(2 \cos a)\left(\frac{\sin x}{x}\right)
\end{aligned}
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0}(2 \cos a)\left(\frac{\sin x}{x}\right) \\
& =(2 \cos a) \lim _{x \rightarrow a}\left(\frac{\sin x}{x}\right) \quad \text { (By Corollary 1.2) } \\
& =(2 \cos a) \times 1 \\
& =2 \cos a
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
f(x) & =\frac{\tan x-\tan a}{x-a} \\
& =\frac{\sin x \cos a-\cos x \sin a}{(x-a) \cos x \cos a} \\
& =\left(\frac{\sin (x-a)}{x-a}\right) \frac{1}{\cos x \cos a}
\end{aligned}
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow a} f(x) & =\lim _{x \rightarrow a}\left(\frac{\sin (x-a)}{x-a}\right) \times \lim _{x \rightarrow a}\left(\frac{1}{\cos x \cos a}\right) \\
& =1 \times \frac{1}{\cos a \cos a} \\
& =\sec ^2 a
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin 2 x}{x \cos x} & =\lim _{x \rightarrow 0} 2\left(\frac{\sin 2 x}{2 x}\right)\left(\frac{1}{\cos x}\right) \\
& =2 \times 1 \times \frac{1}{1} \\
& =2
\end{aligned}
\)

Hint

Let \(0<\delta<1\).
\(
0<\frac{\sin x}{x}<1 \quad \forall x \in(-\delta, \delta)
\)
Therefore
\(
\left[\frac{\sin x}{x}\right]=0 \quad \forall x \in(-\delta, \delta)
\)
Hence \(\lim _{x \rightarrow 0}[f(x)]=0\).

Hint

We have
\(
\frac{x+\sin x}{x-\cos x}=\frac{1+\frac{\sin x}{x}}{1-\frac{\cos x}{x}}
\)
Put \(y=1 / x\) so that \(y \rightarrow 0\) as \(x \rightarrow \infty\). Now
\(
y \sin \frac{1}{y} \rightarrow 0 \text { and } y \cos \frac{1}{y} \rightarrow 0
\)
as \(y \rightarrow 0\), Therefore
\(
\lim _{x \rightarrow \infty} f(x)=\sqrt{\frac{1}{1}}=1
\)

Hint

Put \(y=6 x-\pi\) so that \(y \rightarrow 0\) as \(x \rightarrow \pi / 6\). Also \(x=(\pi+y) / 6\). Therefore
\(
\begin{aligned}
f(x) & =\frac{2-\sqrt{3} \cos \left(\frac{\pi}{6}+\frac{y}{6}\right)-\sin \left(\frac{\pi}{6}+\frac{y}{6}\right)}{y^2} \\
& =\frac{2-\sqrt{3}\left[\frac{\sqrt{3}}{2} \cos \frac{y}{6}-\frac{1}{2} \sin \frac{y}{6}\right]-\left[\frac{1}{2} \cos \frac{y}{6}+\frac{\sqrt{3}}{2} \sin \frac{y}{6}\right]}{y^2} \\
& =\frac{2-2 \cos (y / 6)}{y^2} \\
& =\frac{4 \sin ^2(y / 12)}{y^2} \\
& =4\left[\frac{\sin (y / 12)}{y / 12}\right]^2 \times \frac{1}{144} \\
& =\frac{1}{36}\left(\frac{\sin \theta}{\theta}\right)^2
\end{aligned}
\)
where \(\theta=y / 12 \rightarrow 0\) as \(y \rightarrow 0\). Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\frac{1}{36} \lim _{\theta \rightarrow 0}\left(\frac{\sin \theta}{\theta}\right)^2 \\
& =\frac{1}{36} \times 1^2=\frac{1}{36}
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x} & =\frac{2-(1+\cos x)}{\sin ^2 x} \times \frac{1}{\sqrt{2}+\sqrt{1+\cos x}} \\
& =\left(\frac{2 \sin ^2 \frac{x}{2}}{4 \sin ^2 \frac{x}{2} \cos ^2 \frac{x}{2}}\right) \frac{1}{\sqrt{2}+\sqrt{1+\cos x}} \\
& =\frac{1}{2 \cos ^2 \frac{x}{2}} \times \frac{1}{\sqrt{2}+\sqrt{1+\cos x}}
\end{aligned}
\)
Therefore
\(
\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}=\frac{1}{2} \times \frac{1}{\sqrt{2}+\sqrt{1+1}}=\frac{1}{4 \sqrt{2}}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi\left(1-\sin ^2 x\right)\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \pi \cdot \frac{\sin \left(\pi \sin ^2 x\right)}{\left(\pi \sin ^2 x\right)} \cdot \frac{\sin ^2 x}{x^2} \\
& =\pi \lim _{x \rightarrow 0}\left(\frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2 \\
& =\pi \times 1 \times 1=\pi
\end{aligned}
\)

Hint

Observe that \([x]=0 \Leftrightarrow 0<x<1\). Therefore if \(0<\delta<1\), then
\(
f(x)= \begin{cases}0 & \text { if } 0<x<\delta \\ \frac{\sin (-1)}{(-1)}=\sin 1 & \text { if }-\delta<x<0\end{cases}
\)
So
\(
\lim _{x \rightarrow 0+0} f(x)=0 \text { and } \lim _{x \rightarrow 0-0} f(x)=\sin 1
\)
Hence \(\lim _{x \rightarrow 0} f(x)\) does not exist.

Hint

We have
\(
\frac{\sqrt{1-\cos 2(x-1)}}{x-1}=\sqrt{2} \frac{|\sin (x-1)|}{x-1}
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 1-0} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} & =\lim _{h \rightarrow 0} \frac{\sqrt{2}|\sin (1-h-1)|}{(1-h)-1} \\
& =\lim _{h \rightarrow 0} \sqrt{2}\left(\frac{-\sinh }{h}\right)=-\sqrt{2} \\
\lim _{x \rightarrow 1+0} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} & =\sqrt{2} \lim _{h \rightarrow 0} \frac{|\sin (1+h-1)|}{1+h-1} \\
& =\sqrt{2} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right) \\
& =\sqrt{2}
\end{aligned}
\)
Left limit \(\neq\) Right limit

Hint

Let
\(
\begin{aligned}
f(x) & =\frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2} \\
& =\frac{2 x \tan x\left(\frac{1}{1-\tan ^2 x}-1\right)}{4 \sin ^4 x} \\
& =\frac{x \tan ^3 x}{2 \sin ^4 x} \times \frac{1}{1-\tan ^2 x} \\
& =\frac{x \sin ^3 x}{2 \cos ^3 x \sin ^4 x} \times \frac{\cos ^2 x}{\cos 2 x} \\
& =\frac{1}{2}\left(\frac{x}{\sin x}\right) \frac{1}{\cos x \cos 2 x}
\end{aligned}
\)
So
\(
\lim _{x \rightarrow 0} f(x)=\frac{1}{2}(1) \times \frac{1}{1}=\frac{1}{2}
\)

Hint

Given relation is
\(
f(x \cdot f(y))=f(x y)+x \dots(1)
\)
Interchanging \(x\) and \(y\) in Eq. (1), we have
\(
f(y \cdot f(x))=f(y x)+y \dots(2)
\)
Again replacing \(x\) with \(f(x)\) in Eq. (1) we get
\(
f(f(x) \cdot f(y))=f(y \cdot f(x))+f(x) \dots(3)
\)
Therefore, Eqs. (1)-(3) imply
\(
f(f(x) \cdot f(y))=f(x y)+y+f(x) \dots(4)
\)
Again interchanging \(x\) and \(y\) in Eq. (4), we have
\(
f(f(y) \cdot f(x))=f(y x)+x+f(y) \dots(5)
\)
Equations (4) and (5) imply
\(
\begin{aligned}
& f(x y)+y+f(x)=f(y x)+x+f(y) \dots(6) \\
\Rightarrow & f(x)-x=f(y)-y \forall x, y \in R ^{+}
\end{aligned}
\)
Suppose
\(
f(x)-x=f(y)-y=\lambda
\)
Substituting \(f(x)=\lambda+x\) in Eq. (1), we have
\(
\begin{aligned}
& x \cdot f(y)+\lambda=(x y+\lambda)+x \\
\Rightarrow & x \cdot f(y)=x y+x
\end{aligned}
\)
Therefore
\(
\begin{aligned}
& x(y+\lambda)=x y+x \quad[\because f(y)=\lambda+y] \\
\Rightarrow & \lambda x=x \\
\Rightarrow & \lambda=1 \quad(\because x>0)
\end{aligned}
\)
So \(f(x)=x+\lambda=x+1\)
Hence
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{(f(x))^{1 / 3}-1}{(f(x))^{1 / 2}-1} & =\lim _{x \rightarrow 0} \frac{(1+x)^{1 / 3}-1}{(1+x)^{1 / 2}-1} \\
& =\lim _{x \rightarrow 0}\left(\frac{(1+x)^{1 / 3}-1}{1+x-1}\right) \cdot\left(\frac{1+x-1}{(1+x)^{1 / 2}-1}\right) \\
& =\frac{1 / 3}{1 / 2}=\frac{2}{3}
\end{aligned}
\)

Hint

If \(0<\delta<1\), then \(f(x)=\sin x\) for \(x \in(-\delta, \delta)\), \(x \neq 0\). Therefore
\(
\lim _{x \rightarrow 0} g(f(x))=\lim _{x \rightarrow 0} g(\sin x)=\lim _{x \rightarrow 0}\left(\sin ^2 x+1\right)=1
\)

Hint

Put \(1-x=\theta\) so that \(\theta \rightarrow 0\) as \(x \rightarrow 1\). Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2} & =\lim _{\theta \rightarrow 0} \theta \tan \frac{\pi}{2}(1-\theta) \\
& =\lim _{\theta \rightarrow 0} \theta \cot \left(\frac{\pi \theta}{2}\right) \\
& =\lim _{\theta \rightarrow 0} \frac{\theta \cos \left(\frac{\pi}{2} \theta\right)}{\sin \left(\frac{\pi}{2} \theta\right)} \\
& =\lim _{\theta \rightarrow 0} \frac{2}{\pi} \cdot\left(\frac{\frac{\pi}{2} \theta}{\sin \left(\frac{\pi}{2} \theta\right)}\right) \cos \left(\frac{\pi}{2} \theta\right) \\
& =\frac{2}{\pi} \times 1 \times 1=\frac{2}{\pi}
\end{aligned}
\)

Hint

We have
\(
\begin{gathered}
\lim _{x \rightarrow 1-0} f(x)=\lim _{h \rightarrow 0}\left[\frac{(1-h)^2}{4}-\frac{3}{2}(1-h)+\frac{13}{4}\right]=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2 \\
\lim _{x \rightarrow 1+0} f(x)=\lim _{h \rightarrow 0}|1+h-3|=2
\end{gathered}
\)
Therefore
\(
\lim _{x \rightarrow 1} f(x)=2
\)

Hint

We can see that (Fig. below)

\(
f(x)= \begin{cases}x & \text { if } x \leq 0 \\ x^2 & \text { if } 0<x \leq 1 \\ x & \text { if } x>1\end{cases}
\)
Now
\(
\begin{aligned}
\lim _{h \rightarrow 0-0} \frac{f(1+h)-f(1)}{-h} & =\lim _{h \rightarrow 0} \frac{(1-h)^2-1}{-h} \\
& =\lim _{h \rightarrow 0} \frac{-2 h+h^2}{-h}=2
\end{aligned}
\)
Now
\(
\lim _{h \rightarrow 0+0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{(1+0+h)-1}{0+h}=\lim _{h \rightarrow 0}\left(\frac{h}{h}\right)=1
\)
Therefore the required limit does not exist.

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3} & =\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{x^3 \cos x} \\
& =\lim _{x \rightarrow 0} \frac{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(2 \sin ^2 \frac{x}{2}\right)}{x^3 \cos x} \\
& =\lim _{x \rightarrow 0} 4\left(\frac{\sin \frac{x}{2}}{\left(\frac{x}{2}\right)}\right) \cdot\left(\frac{\cos \frac{x}{2}}{8 \cos x}\right) \\
& =\frac{1}{2} \times 1^3 \times 1=\frac{1}{2}
\end{aligned}
\)

Hint

Let
\(
f(x)=\frac{x+1}{(17+x)^{1 / 4}-2}
\)
Put \((17+x)^{1 / 4}=y\) so that \(y \rightarrow 2\) as \(x \rightarrow-1\). Therefore
\(
f(x)=\frac{y^4-16}{y-2}=\frac{y^4-2^4}{y-2}
\)
So
\(
\lim _{x \rightarrow-1} f(x)=\lim _{y \rightarrow 2}\left(\frac{y^4-2^4}{y-2}\right)=4 \times 2^3=32
\)

Hint

Put \(x=z^{m n}\) so that \(x \rightarrow-1 \Rightarrow z \rightarrow-1\) and \(x^{1 / n}=z^m, x^{1 / m}=z^n\). Therefore
\(
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{1+\sqrt[n]{x}}{1+\sqrt[m]{x}}=\lim _{z-1} \frac{z^m+1}{z^n+1} \\
&=\lim _{z \rightarrow-1}\left(\frac{z^m-(-1)^m}{z+1} \cdot \frac{z+1}{z^n-(-1)^n}\right) \\
&(\because m, n \text { are odd }) \\
&=m(-1)^{m-1} \cdot \frac{1}{n(-1)^{n-1}} \\
&=\frac{m}{n}(\because m, n \text { are odd })
\end{aligned}
\)

Hint

Given limit is
\(
\lim _{x \rightarrow \frac{\pi}{6}} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)}=\lim _{x \rightarrow \frac{\pi}{6}}\left(\frac{\sin x+1}{\sin x-1}\right)
\)
\(
=\frac{(1 / 2)+1}{(1 / 2)-1}=-3
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0}\left(\frac{1-\cos m x}{1-\cos n x}\right) & =\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{m x}{2}\right)}{2 \sin ^2\left(\frac{n x}{2}\right)} \\
& =\lim _{x \rightarrow 0}\left(\frac{\sin \left(\frac{m x}{2}\right)}{m x / 2} \cdot \frac{m x}{2}\right)^2\left(\frac{n x / 2}{\sin \left(\frac{n x}{2}\right)} \cdot \frac{1}{n x / 2}\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right)^2 \cdot \lim _{x \rightarrow 0}\left(\frac{\frac{n x}{2}}{\sin \frac{n x}{2}}\right)^2 \cdot \frac{m^2}{n^2} \\
& =1^2 \times 1^2 \times \frac{m^2}{n^2}=\frac{m^2}{n^2}
\end{aligned}
\)

Hint

Put \(\theta=\operatorname{Sin}^{-1} x\) so that \(\theta \rightarrow 0\) as \(x \rightarrow 0\) and \(x=\sin \theta\). Therefore
\(
\lim _{x \rightarrow 0}\left(\frac{\operatorname{Sin}^{-1} x}{x}\right)=\lim _{\theta \rightarrow 0}\left(\frac{\theta}{\sin \theta}\right)=1
\)

Hint

Let
\(
f(x)=\frac{1+\cos \pi x}{\tan ^2 \pi x}=\frac{(1+\cos \pi x) \cos ^2 \pi x}{\sin ^2 \pi x}
\)
\(
=\frac{(1+\cos \pi x) \cos ^2 \pi x}{1-\cos ^2 \pi x}=\frac{\cos ^2 \pi x}{1-\cos \pi x}
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 1} f(x) & =\lim _{x \rightarrow 1} \frac{\cos ^2 \pi x}{1-\cos \pi x} \\
& =\frac{\cos ^2(\pi)}{1-\cos \pi} \\
& =\frac{(-1)^2}{1-(-1)}=\frac{1}{2}
\end{aligned}
\)

Hint

We have
\(
\frac{x^{-1 / 3}-1}{x^{-2 / 3}-1}=\frac{x^{-1 / 3}-1}{x-1} \cdot \frac{x-1}{x^{-2 / 3}-1}
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 1}\left(\frac{x^{-1 / 3}-1}{x^{-2 / 3}-1}\right) & =\lim _{x \rightarrow 1}\left(\frac{x^{-1 / 3}-1}{x-1}\right) \cdot \lim _{x \rightarrow 1}\left(\frac{x-1}{x^{-2 / 3}-1}\right) \\
& =\frac{-\frac{1}{3}(1)^{(-1 / 3)-1}}{-\frac{2}{3}(1)^{(-2 / 3)-1}}=\frac{1}{2}
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x) & =\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1-\sin x}{\cos x}\right) \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)} \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}
\end{aligned}
\)
\(
\begin{gathered}
=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}{\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)} \\
=\frac{\sin \left(\frac{\pi}{4}-\frac{\pi}{4}\right)}{\cos \left(\frac{\pi}{4}-\frac{\pi}{4}\right)}=0
\end{gathered}
\)
Alternate:
\(
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x) & =\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1-\sin x}{\cos x}\right) \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\sqrt{1-\sin ^2 x}} \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \sqrt{\frac{1-\sin x}{1+\sin x}} \\
& =\sqrt{\frac{1-1}{1+1}}=0
\end{aligned}
\)

Hint

\(f(x)\) is defined in a deleted neighbourhood 1.
Also
\(
f(x)=\frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)}=\frac{(2 x-3)}{(2 x+3)(\sqrt{x}+1)}
\)
Therefore
\(
\lim _{x \rightarrow 1} f(x)=\frac{2-3}{(2+3)(1+1)}=-\frac{1}{10}
\)

Hint

\(g(x)\) is defined in \(a\) neighbourhood of \(\pi / 4\) except at \(\pi / 4\). Now,
\(
\begin{aligned}
g(x) & =\frac{\tan x(\tan x+1)(\tan x-1)}{(1 / \sqrt{2})(\cos x-\sin x)} \\
& =\frac{\sqrt{2} \tan x(\tan x+1)(\sin x-\cos x)}{\cos x(\cos x-\sin x)} \\
& =\frac{-\sqrt{2} \tan x(\tan x+1)}{\cos x}
\end{aligned}
\)
Therefore
\(
\lim _{x \rightarrow \frac{\pi}{4}} g(x)=\frac{-\sqrt{2}(1)(1+1)}{1 / \sqrt{2}}=-2(2)=-4
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow+\infty} x\left(\sqrt{x^2+1}-x\right) & =\lim _{x \rightarrow+\infty} \frac{x\left(x^2+1-x^2\right)}{\sqrt{x^2+1}+x} \\
& =\lim _{x \rightarrow+\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}+1} \\
& =\frac{1}{1+1}=\frac{1}{2}
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow-\infty}\left(\sqrt{2 x^2-3}-5 x\right) & =\lim _{x \rightarrow-\infty}\left(\frac{2 x^2-3-25 x^2}{\sqrt{2 x^2-3}+5 x}\right) \\
& =-\lim _{x \rightarrow-\infty}\left(\frac{23 x^2+3}{\sqrt{2 x^2-3}+5 x}\right) \\
& =-\lim _{x \rightarrow-\infty}\left(\frac{23 x+\frac{3}{x}}{\sqrt{2-\frac{3}{x^2}+5 x}}\right)
\end{aligned}
\)
\(
=\frac{-(-\infty)}{\sqrt{2}+5}=+\infty
\)

Hint

Let
\(
f(x)=\frac{\cos x}{(1-\sin x)^{2 / 3}}
\)
Put \(\theta=(\pi / 2)-x\) so that \(\theta \rightarrow 0\) as \(x \rightarrow \pi / 2\). Now
\(
\begin{aligned}
f(x) & =\frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\left(1-\sin \left(\frac{\pi}{2}-\theta\right)\right)^{2 / 3}} \\
& =\frac{\sin \theta}{(1-\cos \theta)^{2 / 3}} \\
& =\frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2^{2 / 3} \cdot \sin ^{4 / 3}\left(\frac{\theta}{2}\right)} \\
& =\frac{2^{1 / 3} \cos \theta}{\sin ^{1 / 3}\left(\frac{\theta}{2}\right)} \rightarrow \infty \quad \text { as } \theta \rightarrow 0
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
f(x) & =\frac{x^3\left\{x^2+\sqrt{x^4+1}-2 x^2\right\}}{\sqrt{x^2+\sqrt{x^4+1}}+x \sqrt{2}} \\
& =\frac{x^3\left\{\sqrt{x^4+1}-x^2\right\}}{\sqrt{x^2+\sqrt{x^4+1}}+x \sqrt{2}} \\
& =\frac{x^3\left(x^4+1-x^4\right)}{\left[\sqrt{x^2+\sqrt{x^4+1}}+x \sqrt{2}\right]\left[\sqrt{x^4+1}+x^2\right]} \\
& =\frac{x^3}{\left[\sqrt{x^2 \sqrt{x^4+1}}+x \sqrt{2}\right]\left[\sqrt{x^4+1}+x^2\right]} \\
& =\frac{1}{\left[\sqrt{1+\sqrt{1+\frac{1}{x^4}}}+\sqrt{2}\right]\left[\sqrt{1+\frac{1}{x^4}}+1\right]}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{(\sqrt{1+\sqrt{1}}+\sqrt{2})(\sqrt{1}+1)} \\
& =\frac{1}{2 \sqrt{2}(2)}=\frac{1}{4 \sqrt{2}}
\end{aligned}
\)

Hint

In a neighbourhood of zero,
\(
\operatorname{Sin}^{-1}\left(\frac{2 x}{1+x^2}\right)=2 \operatorname{Tan}^{-1} x
\)
Therefore
\(
\begin{aligned}
\lim _{x \rightarrow 0}\left(\frac{1}{x} \operatorname{Sin}^{-1} x\right) & =\lim _{x \rightarrow 0}\left(\frac{2 \operatorname{Tan}^{-1} x}{x}\right) \\
& =2 \lim _{\theta \rightarrow 0}\left(\frac{\theta}{\tan \theta}\right) \text { where } \theta=\operatorname{Tan}^{-1} x \\
& =2 \times 1=2
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow+\infty}\left(2^{x-1} \tan \left(\frac{3}{2^x}\right)\right) & =\lim _{x \rightarrow+\infty}\left(\frac{1}{2} \frac{\tan \left(\frac{3}{2^x}\right)}{3 / 2^x} \cdot 3\right) \\
& =\frac{3}{2} \lim _{\theta \rightarrow 0}\left(\frac{\tan \theta}{\theta}\right) \quad\left(\text { where } \theta=\frac{3}{2^x}\right) \\
& =\frac{3}{2} \times 1=\frac{3}{2}
\end{aligned}
\)

Hint

We have
\(
\begin{gathered}
\lim _{x \rightarrow 0} \frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}=\lim _{x \rightarrow 0} a^{\sin x}\left(\frac{a^{\tan x-\sin x}-1}{\tan x-\sin x}\right) \\
=\lim _{x \rightarrow 0}\left(a^{\sin x}\right) \times \lim _{t \rightarrow 0}\left(\frac{a^t-1}{t}\right) \quad(\text { where } t=\tan x-\sin x) \\
=a^0 \times \log _e a=\log _e a 
\end{gathered}
\)

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow 0}\left(\frac{\log (6+x)-\log (6-x)}{x}\right)= & \lim _{x \rightarrow 0} \frac{\frac{1}{6} \log \left(1+\frac{x}{6}\right)}{x / 6} \\
& +\lim _{x \rightarrow 0} \frac{\frac{1}{6} \log \left(1-\frac{x}{6}\right)}{-x / 6} \\
& =\frac{1}{6} \times 1+\frac{1}{6} \times 1=\frac{1}{3}
\end{aligned}
\)

Hint

The given limit can be written as
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(4^x-1\right)(\sqrt{2 \cos x+7}+3)}{2 \cos x-2} \\
& =\lim _{x \rightarrow 0} \frac{\left(\frac{3^x-1}{x}\right)\left(\frac{4^x-1}{x}\right)(\sqrt{2 \cos x+7}+3) x^2}{-4 \sin ^2 \frac{x}{2}} \\
& =\lim _{x \rightarrow 0} \frac{\left(\frac{3^x-1}{x}\right)\left(\frac{4^x-1}{x}\right)(\sqrt{2 \cos x+7}+3)}{-[\sin (x / 2) /(x / 2)]^2} \\
& =-\log _e 3 \times \log _e 4 \times 6
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
\text { Given limit } & =\lim _{x \rightarrow 0} \frac{\left(5^x-1\right)\left(2^x-1\right)}{x \sin x} \\
& =\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)\left(\frac{2^x-1}{x}\right)\left(\frac{x}{\sin x}\right) \\
& =\log _e 5 \cdot \log _e 2 \cdot 1 \\
& =\log _e 5 \cdot \log _e 2
\end{aligned}
\)

Hint

We have
\(
\frac{G(x)-G(1)}{x-1}=-\frac{\sqrt{25-x^2}-(-\sqrt{24})}{x-1}
\)
\(
\begin{aligned}
& =-\frac{\left[\sqrt{25-x^2}-\sqrt{24}\right]}{x-1} \\
& =-\frac{\left[25-x^2-24\right]}{x-1}-\frac{1}{\left(\sqrt{25-x^2}+\sqrt{24}\right)} \\
& =\frac{x+1}{\sqrt{25-x^2}+\sqrt{24}}
\end{aligned}
\)
Therefore
\(
\lim _{x \rightarrow 1}\left(\frac{G(x)-G(1)}{x-1}\right)=\frac{1+1}{\sqrt{24}+\sqrt{24}}=\frac{1}{\sqrt{24}}=\frac{1}{2 \sqrt{6}}
\)

Hint

We have
\(
\begin{aligned}
\left(\frac{x+6}{x+1}\right)^{x+4} & =\frac{\left(1+\frac{6}{x}\right)^x}{\left(1+\frac{1}{x}\right)^x} \cdot\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}\right)^4 \\
& =\frac{\left[\left(1+\frac{6}{x}\right)^{x / 6}\right]^6}{\left(1+\frac{1}{x}\right)^x} \cdot\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}\right)^4
\end{aligned}
\)
Therefore
\(
\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=\frac{e^6}{e} \cdot\left(\frac{1+0}{1+0}\right)=e^5
\)

Hint

We have
\(
\left(\frac{1+6 x^2}{1+3 x^2}\right)^{1 / x^2}=\frac{\left[\left(1+6 x^2\right)^{1 / 6 x^2}\right]^6}{\left[\left(1+3 x^2\right)^{1 / 3 x^2}\right]^3}
\)
\(
\lim _{x \rightarrow 0}\left(\frac{1+6 x^2}{1+3 x^2}\right)^{1 / x^2}=\frac{e^6}{e^3}=e^3
\)

Hint

Let
\(
\begin{aligned}
f(x) & =\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{1 / x} \\
& =\left(\frac{1+\tan x}{1-\tan x}\right)^{1 / x} \\
& =\frac{\left[(1+\tan x)^{1 / \tan x}\right]^{\tan x / x}}{\left[(1-\tan x)^{-1 / \tan x}\right]^{-\tan x / x}}
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} f(x)=\frac{e^1}{e^{-1}}=e^2
\)

Hint

We have
\(
(\cos x)^{\cot x}=\left[(1+\cos x-1)^{1 /(\cos x-1)}\right]^{(\cos x-1) \tan x}
\)
Take
\(
f(x)=[1+(\cos x-1)]^{1 /(\cos x-1)} \text { and } g(x)=\frac{\cos x-1}{\tan x}
\)
We know that [by part (1) of Important Formulae]
\(
\lim _{x \rightarrow 0} f(x)=e
\)
Now,
\(
g(x)=\frac{-2 \sin ^2 \frac{x}{2} \cos x}{\sin x}=-\frac{\sin \frac{x}{2} \cos x}{\cos \frac{x}{2}} \rightarrow 0 \text { as } x \rightarrow 0
\)
Therefore, by part (7) of Important Formulae
\(
\lim _{x \rightarrow 0}(f(x))^{g(x)}=e^0=1
\)

Hint

Let
\(
\begin{aligned}
f(x) & =\frac{(1-\sin x)\left(8 x^3-\pi^3\right) \cos x}{(\pi-2 x)^4} \\
& =\frac{(1-\sin x) \cos x(2 x-\pi)\left(4 x^2+2 \pi x+\pi^2\right)}{(2 x-\pi)^4} \\
& =\frac{(1-\sin x) \cos x\left(4 x^2+2 \pi x+\pi^2\right)}{(2 x-\pi)^3}
\end{aligned}
\)
Therefore
\(
\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x) \cos x}{(2 x-\pi)^3} \cdot\left(3 \pi^2\right) \dots(1)
\)
Put \(2 x-\pi=y\) so that \(y \rightarrow 0\) as \(x \rightarrow \pi / 2\). Therefore now
\(
\begin{aligned}
\frac{(1-\sin x) \cos x}{(2 x-\pi)^3} & =\frac{\left[1-\sin \left(\frac{\pi+y}{2}\right)\right] \cos \left(\frac{\pi+y}{2}\right)}{y^3} \\
& =\frac{\left(1-\cos \frac{y}{2}\right)\left(-\sin \frac{y}{2}\right)}{y^3} \\
& =-\left(\frac{2 \sin ^2 \frac{y}{4}}{y^2}\right)\left(\frac{\sin \frac{y}{2}}{y}\right) \\
& =-2\left(\frac{\sin ^2 \frac{y}{4}}{y / 4}\right)^2 \cdot \frac{1}{16} \cdot\left(\frac{\sin \frac{y}{2}}{y / 2}\right) \cdot \frac{1}{2} \\
& =-\frac{1}{16}\left(\frac{\sin \frac{y}{4}}{y / 4}\right)^2\left(\frac{\sin \frac{y}{2}}{y / 2}\right) \dots(2)
\end{aligned}
\)
Therefore from Eqs. (1) and (2)
\(
\lim _{x \rightarrow 0} f(y)=\frac{-3 \pi^2}{16} \times 1 \times 1
\)

Hint

Suppose
\(
f(x)=\left(2-\frac{1}{x}\right)^{\tan (\pi x / 2)}=\left(1+\left(1-\frac{1}{x}\right)\right)^{\tan (\pi x / 2)}
\)
Put \(y=1-(1 / x)\) so that \(y \rightarrow 0\) as \(x \rightarrow 1\). Therefore
\(
\begin{aligned}
f(x) & =(1+y)^{\tan (\pi / 2)(1-y)} \\
& =(1+y)^{\cot (\pi y / 2)} \\
& =\left[(1+y)^{1 / y}\right]^{y \cot (\pi y / 2)} \\
& =\left[(1+y)^{1 / y}\right]^{[(\pi y / 2) / \tan (\pi y / 2)] \times(2 / \pi)}
\end{aligned}
\)
\(
\text { Given limit }=e^{1 \times 2 / \pi}=e^{2 / \pi}
\)

Hint

We have
\(
\lim _{x \rightarrow 0} \frac{a e^x-b}{x}=2
\)
Since the denominator \(x \rightarrow 0\) as \(x \rightarrow 0\)
\(
\lim _{x \rightarrow 0}\left(a e^x-b\right)=0
\)
which implies that \(a-b=0\).
\(
2=\lim _{x \rightarrow 0}\left(\frac{a e^x-a}{x}\right)=\lim _{x \rightarrow 0} a\left(\frac{e^x-1}{x}\right)=a
\)
Hence
\(
a=2=b
\)

Hint

Let
\(
\begin{aligned}
f(x) & =\frac{x^2+1}{x+1}-a x-b \\
& =\frac{\left(x^2+1\right)-(x+1)(a x+b)}{x+1} \\
& =\frac{(1-a) x^2-(a+b) x+1-b}{x+1}
\end{aligned}
\)
Put \(y=1 / x\) so that \(y \rightarrow 0\) as \(x \rightarrow \infty\). Therefore
\(
f(x)=f\left(\frac{1}{y}\right)=\frac{(1-a)-(a+b) y+(1-b) y^2}{y(1+y)}
\)
Now,
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} f(x)=0 \\
\Rightarrow & \lim _{y \rightarrow 0} \frac{(1-a)-(a+b) y+(1-b) y^2}{y(y+1)}=0
\end{aligned}
\)
Since the denominator tends to zero as \(y \rightarrow 0\), the numerator must tend to zero as \(y \rightarrow 0\). Therefore
\(
\begin{gathered}
1-a=0 \quad \text { or } \quad a=1 \\
0=\lim _{y \rightarrow 0} \frac{[-(a+b)+(1-b) y]}{y+1}=-(a+b) \\
a=-b
\end{gathered}
\)
Hence \((a, b)=(1,-1)\).

Hint

We have
\(
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{x^3 \sin \frac{1}{x}+2 x^2}{1+3 x^2} & =\lim _{x \rightarrow \infty} \frac{x \sin \frac{1}{x}+2}{\frac{1}{x^2}+3} \\
& =\frac{1+2}{0+3}\left[\because \lim _{x \rightarrow \infty}\left(x \sin \frac{1}{x}\right)=1\right] \\
& =\frac{3}{3}=1
\end{aligned}
\)

Hint

It is known that \(K^{\sec ^2 x} \leq n^{\sec ^2 x}\) for \(K=1,2\), \(3, \ldots, n\). Therefore
\(
\left(1^{\sec ^2 x}+2^{\sec ^2 x}+\cdots+n^{\sec ^2 x}\right)^{\cos ^2 x} \leq\left(n \cdot n^{\sec ^2 x}\right)^{\cos ^2 x}=n \cdot n^{\cos ^2 x}
\)
Again
\(
\left(1^{\sec ^2 x}+2^{\sec ^2 x}+\cdots+n^{\sec ^2 x}\right)^{\cos ^2 x} \geq\left(n^{\sec ^2 x}\right)^{\cos ^2 x}=n
\)
Therefore
\(
\begin{aligned}
n & =\lim _{x \rightarrow \frac{\pi}{2}}(n) \leq \lim _{x \rightarrow \frac{\pi}{2}}\left(1^{\sec ^2 x}+2^{\sec ^2 x}+\cdots+n^{\sec ^2 x}\right)^{\cos ^2 x} \\
& \leq \lim _{x \rightarrow \frac{\pi}{2}}\left(n \cdot n^{\cos ^2 x}\right)=n
\end{aligned}
\)
By squeezing theorem, we have
\(
\lim _{x \rightarrow \frac{\pi}{2}}\left(1^{\sec ^2 x}+2^{\sec ^2 x}+\cdots+n^{\sec ^2 x}\right)^{\cos ^2 x}=n
\)

Try it out:
\(
\lim _{x \rightarrow 0}\left(1^{\operatorname{cosec}^2 x}+2^{\operatorname{cosec}^2 x}+\cdots+n^{\operatorname{cosec}^2 x}\right)^{\sin ^2 x}=n
\)

Hint

We have
\(
x_n=\frac{n(n+1)}{2 n^2}=\frac{1}{2}\left(1+\frac{1}{n}\right) \rightarrow \frac{1}{2} \quad \text { as } n \rightarrow \infty
\)

Hint

We have
\(
\begin{aligned}
& \quad y_n=\frac{n(n+1)(2 n+1)}{6 n^3}=\frac{1}{6}(1)\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) \rightarrow \frac{2}{6}=\frac{1}{3} \\
& \text { as } n \rightarrow \infty .
\end{aligned}
\)

Hint

We have
\(
z_n=\frac{n^2(n+1)^2}{4 n^4}=\frac{1}{4}\left(1+\frac{1}{n}\right)^2 \rightarrow \frac{1}{4} \quad \text { as } n \rightarrow \infty
\)

Hint

By the definition of the integral part
\(
\begin{gathered}
x-1<[x] \leq x \\
2 x-1<[2 x] \leq 2 x \\
3 x-1<[3 x] \leq 3 x \\
\cdots \cdots \cdots \cdots \\
\cdots \cdots \cdots \cdots \\
n x-1<[n x] \leq n x
\end{gathered}
\)
Adding all the inequalities we have
\(
\begin{aligned}
x(1+2+3+\cdots+n)-n & <[1 x]+[2 x]+\cdots+[n x] \\
& \leq x(1+2+\cdots+n) \\
\frac{x}{2} n(n+1)-n<[1 x]+[2 x]+\cdots+[n x] & \leq \frac{x n(n+1)}{2}
\end{aligned}
\)
Dividing throughout by \(n^2\), we get
\(
\frac{x}{2}\left(1+\frac{1}{n}\right)-\frac{1}{n}<\frac{[1 x]+[2 x]+\cdots+[n x]}{n^2} \leq \frac{x}{2}\left(1+\frac{1}{n}\right)
\)
Taking limit as \(n \rightarrow \infty\) and using squeezing theorem we have
\(
\lim _{n \rightarrow \infty} \frac{[1 x]+[2 x]+\cdots+[n x]}{n^2}=\frac{x}{2}
\)

Note: Try it out On similar lines (i.e., by using the concept of \([x])\) we can show that
1. \(\lim _{n \rightarrow \infty} \frac{\left[1^2 x\right]+\left[2^2 x\right]+\cdots+\left[n^2 x\right]}{n^3}=\frac{x}{3}\)
2. \(\lim _{n \rightarrow \infty} \frac{\left[1^3 x\right]+\left[2^3 x\right]+\cdots+\left[n^3 x\right]}{n^4}=\frac{x}{4}\)

Hint

We know that \([x+y]=[x]+[y]\) if one of \(x\) or \(y\) is an integer. Therefore
\(
\begin{gathered}
{[x+1]=[x]+1} \\
{[2 x+2]=[2 x]+2} \\
{[3 x+2]=[3 x]+3} \\
\cdots \cdots \cdots \cdots \cdots \cdots \\
\cdots \cdots \cdots \cdots \cdots \cdots \\
{[n x+n]=[n x]+n}
\end{gathered}
\)
Adding all the equations and dividing by \(n^2\), we get
\(
\begin{aligned}
\lim _{n \rightarrow \infty} a_n & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left\{[x]+[2 x]+\cdots+[n x]+\frac{1}{2} n(n+1)\right\} \\
& =\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+\cdots+[n x]}{n^2}+\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2} \\
& =\frac{x}{2}+\frac{1}{2} \quad \text { (by Problem 98) }
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
& x_1=1, x_2=\frac{4+3}{3+2}=\frac{7}{5} \\
& x_3=\frac{4+3 x_2}{3+2 x_2}=\frac{4+3\left(\frac{7}{5}\right)}{3+2\left(\frac{7}{5}\right)}=\frac{41}{29}>x_2
\end{aligned}
\)
We can easily verify that \(x_n<x_{n+1}\) and hence \(\left\{x_n\right\}\) is strictly increasing sequence of positive terms. Let \(\lim _{n \rightarrow \infty} x_n=l\). Therefore
\(
\begin{aligned}
l & =\lim _{n \rightarrow \infty} x_{n+1} \\
& =\lim _{n \rightarrow \infty}\left(\frac{4+3 x_n}{3+2 x_n}\right) \\
& =\frac{4+3 \lim _{n \rightarrow \infty} x_n}{3+2 \lim _{n \rightarrow \infty} x_n}
\end{aligned}
\)
\(
=\frac{4+3 l}{3+2 l}
\)
Hence \(3 l+2 l^2=4+3 l\)
\(
l^2=2 \Rightarrow l=\sqrt{2} \quad\left(\because x_n>0 \quad \forall n\right)
\)

Hint

\(\left\{x_n\right\}\) is an increasing sequence. Define
\(
\theta_n=\operatorname{Cot}^{-1}\left(x_n\right) \text { or } x_n=\cot \theta_n
\)
Now
\(
x_0=0=\cot \frac{\pi}{2}, x_1=1=\cot \frac{\pi}{2^2}
\)
and in general
\(
\begin{aligned}
x_{n+1} & =\cot \theta_n+\operatorname{cosec} \theta_n \\
& =\frac{1+\cos \theta_n}{\sin \theta_n} \\
& =\cot \left(\frac{\theta_n}{2}\right) \\
& =\cot \left(\frac{\pi}{4 \times 2^n}\right) \\
\left(\because x_p=1=\cot \left(\frac{\pi}{4}\right),\right. & \left.x_2=\cot \left(\frac{\pi}{8}\right)=\cot \left(\frac{\pi}{4 \times 2}\right), \text { etc. }\right)
\end{aligned}
\)
Therefore
\(
\begin{aligned}
\frac{x_n}{2^{n-1}} & =\frac{\cot \left(\frac{\pi}{4 \times 2^{n-1}}\right)}{2^{n-1}} \\
& =\frac{1}{2^{n-1} \tan \left(\frac{\pi}{4 \times 2^{n-1}}\right)}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{\frac{\tan \left(\frac{\pi}{4 \times 2^{n-1}}\right)}{\left(\frac{\pi}{4 \times 2^{n-1}}\right)} \cdot \frac{\pi}{4}} \\
& =\frac{4}{\pi} \frac{1}{\left(\frac{\tan x}{x}\right)}
\end{aligned}
\)
where \(x=\frac{\pi}{4 \times 2^{n-1}} \rightarrow 0\) as \(n \rightarrow \infty\). Therefore
\(
\lim _{n \rightarrow \infty} x_n=\frac{4}{\pi} \cdot 1=\frac{4}{\pi}
\)

Hint

From the given relation we have
\(
\begin{aligned}
& 1 \cdot 2 \cdot a_2=0-(1-2) a_0=1 \\
& \Rightarrow a_2=\frac{1}{2}=\frac{1}{2!}
\end{aligned}
\)
\(
\text { and } \quad 6 \cdot a_3=2 a_2-0 \cdot a_1=2\left(\frac{1}{2}\right)=1
\)
\(
\Rightarrow a_3=\frac{1}{6}=\frac{1}{3!}
\)
Assume that \(a_k=1 / {k}\) for all \(k=2,3, \ldots, m\). Now
\(
\begin{aligned}
m(m+1) \cdot a_{m+1} & =m(m-1) a_m-(m-2) a_{m-1} \\
& =\frac{m(m-1)}{{m!}}-\frac{m-2}{(m-1)!} \\
& =\frac{1}{(m-2)!}-\frac{m-2}{(m-1)!} \\
& =\frac{(m-1)-(m-2)}{(m-1)!}=\frac{1}{(m-1)!}
\end{aligned}
\)
Therefore
\(
a_{m+1}=\frac{1}{m(m+1)(m-1)!}=\frac{1}{(m+1)!}
\)
Hence by complete induction, \(a_k=1 / k\) for \(k \geq 2\).
Therefore
\(
\lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty}\left(\frac{1}{{n!}}\right)=0
\)

Hint

We have
\(
\begin{aligned}
\frac{s_n(x)}{x-1}= & \frac{x}{x^2-1}+\frac{x^2}{x^4-1}+\frac{x^4}{x^{16}-1}+\cdots+\frac{x^{2^{n-1}}}{x^{2^n}-1} \\
= & \left(\frac{1}{x-1}-\frac{1}{x^2-1}\right)+\left(\frac{1}{x^2-1}-\frac{1}{x^4-1}\right)+\cdots \\
& +\left(\frac{1}{x^{2^{n-1}}-1}-\frac{1}{x^{2^n}-1}\right) \\
= & \frac{1}{x-1}-\frac{1}{x^{2^n}-1}
\end{aligned}
\)
Therefore
\(
s_n(x)=1-\frac{x-1}{x^{2^n}-1}
\)
Since \(x>1\), we have \(\lim _{n \rightarrow \infty}\left(x^{2^n}-1\right)=+\infty\) so that
\(
\frac{1}{x^{2^n}-1} \rightarrow 0 \quad \text { as } n \rightarrow \infty
\)
Therefore
\(
\lim _{n \rightarrow \infty} s_n(x)=1-0=1
\)

Hint

We have
\(
\begin{aligned}
P_n & =\left(\frac{a_1+1}{a_1}\right)\left(\frac{a_2+1}{a_2}\right) \cdots\left(\frac{a_n+1}{a_n}\right) \\
& =\left(\frac{a_2}{2 a_1}\right)\left(\frac{a_3}{3 a_2}\right) \cdots\left(\frac{a_n+1}{(n+1) a_n}\right)\left(\because \frac{a_k}{k}=a_{k-1}+1 \text { for } k \geq 2\right)
\end{aligned}
\)
\(
=\frac{a_{n+1}}{(n+1)!} \dots(1)
\)
Now
\(
\begin{aligned}
& \frac{a_n+1}{(n+1)!}=\frac{(n+1)\left(a_n+1\right)}{(n+1)!} \\
\Rightarrow & \frac{a_n+1}{(n+1)!}-\frac{a_n}{{n!}}=\frac{1}{{n!}}
\end{aligned}
\)
Therefore
\(
\begin{aligned}
& \frac{a_2}{2!}-\frac{a_1}{1!}=\frac{1}{1!} \\
& \frac{a_3}{{3!}}-\frac{a_2}{2!}=\frac{1}{{2!}} \\
& \frac{a_{n+1}}{(n+1)!}-\frac{a_n}{n!}=\frac{1}{n!} \\
&
\end{aligned}
\)
Adding all the above equations we get
\(
\frac{a_{n+1}}{(n+1)!}-\frac{a_1}{1!}=\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}
\)
From Eq. (1) we have
\(
P_n=1+\frac{1}{1!}+\frac{1}{2!}+ \dots + \frac{1}{n!} \left(\therefore a_1=1\right)
\)
But the number \(e\) is the sum of the infinite series \(\sum_{n=0}^{\infty} \frac{1}{n!}\) (this is to be assumed). Note that \(\sum a_n\) is defined as the limit of the sequence of the partial sums \(\left\{s_n\right\}\) where \(s_n=a_1+a_2+\cdots+a_n\). Thus
\(
\lim _{n \rightarrow \infty} P_n=e
\)

Hint

We have
\(
\begin{gathered}
f^{(2)}(x)=f\left(f^{(1)}(x)\right)=f(f(x))=\frac{f(x)}{2}+1=\frac{x}{2^2}+\frac{1}{2}+1 \\
f^{(3)}(x)=f\left(f^{(2)}(x)\right)=\frac{1}{2}\left(\frac{x}{2^2}+\frac{1}{2}+1\right)+1=\frac{x}{2^3}+\frac{1}{2^2}+\frac{1}{2}+1
\end{gathered}
\)
By induction we can see that
\(
\begin{aligned}
f^{(n)}(x) & =\frac{x}{2^n}+\frac{1}{2^{n-1}}+\frac{1}{2^{n-2}}+\cdots+\frac{1}{2}+1 \\
& =\frac{x}{2^n}+\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}} \\
& =\frac{x}{2^n}+2-\frac{1}{2^{n-1}}
\end{aligned}
\)
Now
\(
\lim _{n \rightarrow \infty} \frac{1}{2^n}=0 \Rightarrow \lim _{n \rightarrow \infty} f^{(n)}(x)=0+2-0=2
\)

Hint

We have
\(
a_{n+1}-a_n=2 n+1
\)
This implies \(d_n\) divides \(2 n+1\) for all \(n=1,2,3, \ldots\). Now
\(
3 a_n+a_{n+1}=4 n^2+2 n+81=2 n(2 n+1)+81
\)
This implies \(d_n\) divides 81 and 81 will be the L.C.M. of \(d_n\) for \(n=1,2,3, \ldots\) so that \(d=81\). Now
\(
\begin{aligned}
d+\frac{d}{4}+\frac{d}{4^2}+\cdots \infty & =\frac{d}{1-\frac{1}{4}} \\
& =(81) \frac{4}{3}=108
\end{aligned}
\)

Hint

\(
\text { Put } m=n^2+\cos n \text { so that } m \rightarrow \infty=\text { as } n \rightarrow \infty \text {. }
\)
Now
\(
\begin{aligned}
\frac{n^2+n}{m} & =\frac{n^2+n}{n^2+\cos n} \\
& =\frac{1+\frac{1}{n}}{1+\frac{\cos n}{n^2}} \rightarrow 1 \text { as } n \rightarrow \infty
\end{aligned}
\)
because
\(
\lim _{n \rightarrow \infty} \frac{1}{n}=0 \text { and } \lim _{n \rightarrow \infty} \frac{\cos n}{n^2}=0
\)
Therefore
\(
\lim _{n \rightarrow \infty}\left(1+\frac{1}{n^2+\cos n}\right)^{n^2+n}=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{m}\right)^m\right]^{\left(n^2+n\right) / m}=e
\)

Hint

We have \(x_n=2+\sqrt{x_{n-1}}\). Suppose \(\lim _{n \rightarrow \infty}=l\). Then
\(
\begin{aligned}
& l=2+\sqrt{l} \\
\Rightarrow & l^2-5 l+4=0 \\
\Rightarrow & (l-1)(l-4)=0
\end{aligned}
\)
\(
\Rightarrow l=1 \text { or } 4
\)
But \(x_n>2\) and \(\left\{x_n\right\}\) is an increasing sequence and hence \(l\) \(\neq 1\). Thus \(l=4\).

Hint

The given limit is
\(
\begin{aligned}
\lim _{n \rightarrow \infty}\left[\log \left(9 n^2\right)-\log \left(n^2+1\right)\right] & =\lim _{n \rightarrow \infty} \log \frac{9 n^2}{n^2+1} \\
& =\lim _{n \rightarrow \infty} \log \left(\frac{9}{1+\frac{1}{n^2}}\right) \\
& =\log \left(\frac{9}{1+0}\right) \\
& =\log 9=2 \log 3
\end{aligned}
\)

Hint

We have
Given limit \(=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)^{n^2}\right]^{\frac{4\left(\sin \frac{\pi}{2 n}\right)}{\pi / 2 n}}=e^{4 \times 1}=e^4\)

Hint

We have
\(
\begin{aligned}
\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^2} & =\sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2 n)^2} \\
& =\frac{\pi^2}{6}-\frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\
& =\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}
\end{aligned}
\)