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If \(f(x)=0\) be a quadratic equation such that \(f(-\pi)=f(\pi) =0\) and \(f\left(\frac{\pi}{2}\right)=-\frac{3 \pi^2}{4}\), then \(\lim _{x \rightarrow-\pi} \frac{f(x)}{\sin (\sin x)}\) is equal to
\(
\begin{aligned}
& \text { (c) Given } f(x)=x^2-\pi^2 \\
& \begin{aligned}
\lim _{x \rightarrow-\pi} \frac{x^2-\pi^2}{\sin (\sin x)} & =\lim _{h \rightarrow 0} \frac{(-\pi+h)^2-\pi^2}{\sin (\sin (-\pi+h))}=\lim _{h \rightarrow 0} \frac{-2 h \pi+h^2}{-\sin (\sin h)} \\
= & \lim _{h \rightarrow 0} \frac{h-2 \pi}{\frac{-\sin (\sin h)}{\sin h} \times \frac{\sin h}{h}}=2 \pi
\end{aligned}
\end{aligned}
\)
If \(f(x)=\frac{\cos x}{(1-\sin x)^{1 / 3}}\), then
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \frac{-\sin t}{(1-\cos t)^{1 / 3}} \\
& =-\lim _{t \rightarrow 0} \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{\left(2 \sin ^2 \frac{t}{2}\right)^{1 / 3}}
\end{aligned}
\)
\(
=-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \frac{t}{2}\left(\sin \frac{t}{2}\right)^{1 / 3}=0
\)
\(\lim _{x \rightarrow-\infty} \frac{x^2 \tan \frac{1}{x}}{\sqrt{8 x^2+7 x+1}}\) is equal to
(a)
\(
\begin{aligned}
& \lim _{x \rightarrow-\infty} \frac{x^2 \tan \frac{1}{x}}{\sqrt{8 x^2+7 x+1}}=\lim _{x \rightarrow-\infty} \frac{x^2 \tan \frac{1}{x}}{-x \sqrt{8+\frac{7}{x}+\frac{1}{x^2}}} \\
& =-\lim _{x \rightarrow-\infty} \frac{\tan \frac{1}{x}}{\frac{1}{x} \sqrt{8+\frac{7}{x}+\frac{1}{x^2}}}=-\frac{1}{2 \sqrt{2}}
\end{aligned}
\)
\(\lim _{x \rightarrow 0}\left[\frac{\sin (\operatorname{sgn}(x))}{(\operatorname{sgn}(x))}\right]\), where \([\cdot]\) denotes the greatest integer function, is equal to
(a)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0+}\left[\frac{\sin (\operatorname{sgn} x)}{\operatorname{sgn}(x)}\right] \\
& =\lim _{x \rightarrow 0^{+}}\left[\frac{\sin 1}{1}\right] \\
& =0 \\
& =\lim _{x \rightarrow 0^{-}}\left[\frac{\sin (\operatorname{sgn} x)}{\operatorname{sgn}(x)}\right] \\
& =\lim _{x \rightarrow 0^{-}}\left[\frac{\sin (-1)}{-1}\right] \\
& =\lim _{x \rightarrow 0^{-}}[\sin 1] \\
& =0
\end{aligned}\\
&\text { Hence, the given limit is } 0 \text {. }
\end{aligned}
\)
\(
\lim _{x \rightarrow \infty} \frac{2+2 x+\sin 2 x}{(2 x+\sin 2 x) e^{\sin x}} \text { is equal to }
\)
(d)
\(
\begin{aligned}
&\text { The given limit is } \lim _{x \rightarrow \infty} \frac{\frac{2}{x}+2+\frac{\sin 2 x}{x}}{\left(2+\frac{\sin 2 x}{x}\right) e^{\sin x}}\\
&=\frac{0+2+0}{(2+0) \times\left(\text { a value between } \frac{1}{e} \text { and } e\right)} \left[\because \lim _{x \rightarrow \infty} \sin x \in(-1,1)\right]
\end{aligned}
\)
Hence limit does not exist
Let \(\lim _{x \rightarrow 0} \frac{[x]^2}{x^2}=l\) and \(\lim _{x \rightarrow 0} \frac{\left[x^2\right]}{x^2}=m\), where \([\cdot]\) denotes greatest integer, then
(b)
\(
\begin{aligned}
& \frac{[x]^2}{x^2}=\left[\begin{array}{c}
0 \text { if } 0<x<1 \\
\frac{1}{x^2} \text { if }-1<x<0
\end{array} \Rightarrow l\right. \text { does not exist } \\
& \frac{\left[x^2\right]}{x^2}=\left[\begin{array}{c}
0 \text { if } 0<x<1 \\
0 \text { if }-1<x<0
\end{array} \Rightarrow m \text { exists and is equal to } 0\right]
\end{aligned}
\)
\(\lim _{x \rightarrow 1} \frac{x \sin (x-[x])}{x-1}\), where \([\cdot]\) denotes the greatest integer function, is equal to
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{x \sin (x-[x])}{x-1} \\
& \text { Now L.H.L. }=\lim _{h \rightarrow 0} \frac{(1-h) \sin (1-h-[1-h])}{(1-h)-1} \\
& =\lim _{h \rightarrow 0} \frac{(1-h) \sin (1-h)}{-h}=-\infty
\end{aligned}
\)
\(
\begin{aligned}
&\text { R.H.L. }=\lim _{h \rightarrow 0} \frac{(1+h) \sin (1+h-[1+h])}{(1+h)-1}=\lim _{h \rightarrow 0} \frac{(1+h) \sin h}{h}=1\\
&\text { Hence, the limit does not exist. }
\end{aligned}
\)
\(\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}\) is equal to
(c)
\(
\begin{aligned}
&\text { The given limit is } \lim _{x \rightarrow 0}\left[(1+\tan x)^{\operatorname{cosec} x} /(1+\sin x)^{\operatorname{cosec} x}\right]\\
&\begin{aligned}
& \left.=\lim _{x \rightarrow 0}\left[(1+\tan x)^{\cot x}\right\}^{\sec x} x /\left\{1 /(1+\sin x)^{\operatorname{cosec} x}\right\}\right] \\
& =e^{\sec 0} \frac{1}{e}=e \frac{1}{e}=1
\end{aligned}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{\sin ^4 x-\sin ^2 x+1}{\cos ^4 x-\cos ^2 x+1}\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\sin ^4 x-\sin ^2 x+1}{\cos ^4 x-\cos ^2 x+1} \\
= & \lim _{x \rightarrow \infty} \frac{\left(1-\cos ^2 x\right)^2-\left(1-\cos ^2 x\right)+1}{\cos ^4 x-\cos ^2 x+1} \\
= & \lim _{x \rightarrow \infty} \frac{\cos ^4 x-\cos ^2 x+1}{\cos ^4 x-\cos ^2 x+1} \\
= & 1
\end{aligned}
\)
\(\lim _{x \rightarrow \infty}\left(\frac{x^3}{3 x^2-4}-\frac{x^2}{3 x+2}\right)\) is equal to
\(
\begin{aligned}
& \text { (d) } \lim _{x \rightarrow \infty}\left(\frac{x^3}{3 x^2-4}-\frac{x^2}{3 x+2}\right) \\
& =\lim _{x \rightarrow \infty} \frac{x^3(3 x+2)-x^2\left(3 x^2-4\right)}{\left(3 x^2-4\right)(3 x+2)} \\
& =\lim _{x \rightarrow \infty} \frac{2 x^3+4 x^2}{9 x^3+6 x^2-12 x-8} \\
& =\lim _{x \rightarrow \infty} \frac{2+\frac{4}{x}}{9+\frac{6}{x}-\frac{12}{x^2}-\frac{8}{x^3}} \\
& =2 / 9
\end{aligned}
\)
If \(f(x)=\frac{2}{x-3}, g(x)=\frac{x-3}{x+4}\) and \(h(x)=-\frac{2(2 x+1)}{x^2+x-12}\), then \(\lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]\) is
\(
\begin{aligned}
& \text { (c) We have } f(x)+g(x)+h(x)=\frac{x^2-4 x+17-4 x-2}{x^2+x-12} \\
& =\frac{x^2-8 x+15}{x^2+x-12}=\frac{(x-3)(x-5)}{(x-3)(x+4)} \\
& \therefore \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]=\lim _{x \rightarrow 3} \frac{(x-3)(x-5)}{(x-3)(x+4)}=-\frac{2}{7}
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}\) is equal to
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \frac{2 x\left(e^x-1\right)}{4 \sin ^2 \frac{x}{2}} \\
& 2 \lim _{x \rightarrow 0}\left[\frac{(x / 2)^2}{\sin ^2 \frac{x}{2}}\right]\left(\frac{e^x-1}{x}\right)=2
\end{aligned}
\)
\(\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)\left(n^2+3 n-1\right)}\) is equal to
(c)
\(
\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)\left(n^2+3 n-1\right)}
\)
\(
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{\left(2+\frac{1}{n}\right)^2}{\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}-\frac{1}{n^2}\right)} \\
& =\frac{(2+0)^2}{(1+0)(1+0+0)}=4
\end{aligned}
\)
The value of \(\lim _{x \rightarrow \pi} \frac{1+\cos ^3 x}{\sin ^2 x}\) is
(d)
\(
\begin{aligned}
&\text { We have } \lim _{x \rightarrow \pi} \frac{1+\cos ^3 x}{\sin ^2 x}\\
&\begin{aligned}
& =\lim _{x \rightarrow \pi} \frac{(1+\cos x)\left(1-\cos x+\cos ^2 x\right)}{(1-\cos x)(1+\cos x)} \\
& =\lim _{x \rightarrow \pi} \frac{1-\cos x+\cos ^2 x}{1-\cos x}=\frac{1+1+1}{1+1}=\frac{3}{2}
\end{aligned}
\end{aligned}
\)
\(\lim _{n \rightarrow \infty} n^2\left(x^{1 / n}-x^{1 /(n+1)}\right), x>0\), is equal to
(c)
\(
\begin{aligned}
& \lim _{n \rightarrow \infty} n^2\left(x^{1 / n}-x^{\frac{1}{n+1}}\right)=\lim _{n \rightarrow \infty} n^2 \cdot x^{\frac{1}{n+1}}\left(x^{\frac{1}{n}-\frac{1}{n+1}}-1\right) \\
& =\lim _{n \rightarrow \infty} x^{\frac{1}{n+1}}\left(x^{\frac{1}{n(n+1)}}-1\right) n^2 \\
& =\lim _{n \rightarrow \infty} x^{\frac{1}{n+1}} \cdot \frac{x^{\frac{1}{n(n+1)}}-1}{\frac{1}{n(n+1)}} \cdot \frac{n^2}{n(n+1)}=1 \cdot \log _e x \cdot 1=\log _e x
\end{aligned}
\)
The value of \(\lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\) is
(a)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{1+\sqrt{2+x}-3}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(x-2)} \text { (Rationalizing) } \\
& =\lim _{x \rightarrow 2} \frac{\sqrt{2+x}-2}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(x-2)} \\
& =\lim _{x \rightarrow 2} \frac{(x-2)}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)(x-2)}
\end{aligned}\\
&\text { (Rationalizing) }\\
&=\frac{1}{(2 \sqrt{3}) 4}=\frac{1}{8 \sqrt{3}}
\end{aligned}
\)
\(
\lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^5}{(2 x+3)^{45}} \text { is equal to }
\)
(c)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^5}{(2 x+3)^{45}} \\
& =\lim _{x \rightarrow \infty} \frac{\left(2+\frac{1}{x}\right)^{40}\left(4-\frac{1}{x}\right)^5}{\left(2+\frac{3}{x}\right)^{45}}
\end{aligned}\\
&\text { (Dividing numerator and denominator by } x^{45} \text { ) }
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{2^{40} 4^5}{2^{45}} \\
& =2^5=32
\end{aligned}
\)
\(\lim _{x \rightarrow \infty}[\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}]\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty}[\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}] \\
& =\lim _{x \rightarrow \infty} \frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}(\text { Rationalizing }) \\
& \lim _{x \rightarrow \infty} \frac{\sqrt{1+x^{-1 / 2}}}{\sqrt{1+\sqrt{x^{-1}+x^{-3 / 2}}}+1}=\frac{1}{2}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}}\) is equal to
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}} \\
& =\lim _{x \rightarrow \infty} \frac{x^{10}\left[\left(1+\frac{1}{x}\right)^{10}+\left(1+\frac{2}{x}\right)^{10}+\cdots+\left(1+\frac{100}{x}\right)^{10}\right]}{x^{10}\left[1+\frac{10^{10}}{x^{10}}\right]} \\
& =100
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{x^a \sin ^b x}{\sin \left(x^c\right)}\), where \(a, b, c \in R \sim\{0\}\), exists and has nonzero value, then
(c)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x^a \sin ^b x}{\sin x^c} \\
& \quad=\lim _{x \rightarrow 0} x^a\left(\frac{\sin x}{x}\right)^b\left(\frac{x^c}{\sin x^c}\right) x^{b-c}=\lim _{x \rightarrow 0} x^{a+b-c}
\end{aligned}\\
&\text { This limit will have non-zero value if } a+b=c \text {. }
\end{aligned}
\)
\(\lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\frac{\pi}{2}\right) \sec x\right]\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\frac{\pi}{2}\right) \sec x\right] \\
& =\lim _{x \rightarrow \pi / 2} \frac{2 x \sin x-\pi}{2 \cos x} \left(\frac{0}{0} \text { form }\right)
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow \pi / 2} \frac{[2 \sin x+2 x \cos x]}{-2 \sin x} \text { (Applying L’Hopital’s rule) } \\
& =-1
\end{aligned}
\)
If \(\lim _{x \rightarrow \infty}\left(\frac{x^3+1}{x^2+1}-(a x+b)\right)=2\), then
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{x^3+1}{x^2+1}-(a x+b)\right)=2 \\
& \Rightarrow \lim _{x \rightarrow \infty} \frac{x^3(1-a)-b x^2-a x+(1-b)}{x^2+1}=2 \\
& \Rightarrow 1-a=0 \text { and }-b=2 \\
& \Rightarrow a=1, b=-2
\end{aligned}
\)
The value of \(\lim _{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}\) is
(c)
\(
\lim _{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \frac{\pi x}{2}} \\
& =e^{\lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}} \\
& =e^{\lim _{x \rightarrow 1}(1-x) \cot \left(\frac{\pi}{2}-\frac{\pi x}{2}\right)} \\
& =e^{\lim _{x \rightarrow 1} \frac{(1-x)}{\tan \left(\frac{\pi}{2}-\frac{\pi x}{2}\right)}} \\
& =e^{\frac{2}{\pi} \lim _{x \rightarrow 1} \frac{\frac{\pi}{2}(1-x)}{\tan \left(\frac{\pi}{2}(1-x)\right)}} \\
& =e^{2 / \pi}
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{\sin x^n}{(\sin x)^m},(m<n)\) is equal to
(b)
\(
\begin{array}{rlrl}
\lim _{x \rightarrow 0} \frac{\sin x^n}{(\sin x)^m} & =\lim _{x \rightarrow 0}\left(\frac{\sin x^n}{x^n}\right)\left(\frac{x^n}{x^m}\right)\left(\frac{x}{\sin x}\right)^m & & \\
=\lim _{x \rightarrow 0} x^{n-m} & =0 & {[\because m<n]}
\end{array}
\)
\(\lim _{x \rightarrow 0} \frac{x^4\left(\cot ^4 x-\cot ^2 x+1\right)}{\left(\tan ^4 x-\tan ^2 x+1\right)}\) is equal to
(a)
\(
\begin{aligned}
& \frac{x^4\left(\cot ^4 x-\cot ^2 x+1\right)}{\left(\tan ^4 x-\tan ^2 x+1\right)} \\
& =\frac{x^4\left(1-\tan ^2 x+\tan ^4 x\right)}{\tan ^4 x\left(\tan ^4 x-\tan ^2 x+1\right)}=\frac{x^4}{\tan ^4 x}, x \neq 0 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{x^4\left(\cot ^4 x-\cot ^2 x+1\right)}{\left(\tan ^4 x-\tan ^2 x+1\right)}=\lim _{x \rightarrow 0} \frac{x^4}{\tan ^4 x}=1
\end{aligned}
\)
\(\lim _{x \rightarrow \infty}\left(\frac{1}{e}-\frac{x}{1+x}\right)^x\) is equal to
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{1}{e}-\frac{x}{1+x}\right)^x=\lim _{x \rightarrow \infty}\left(\frac{1}{e}-\frac{1}{\frac{1}{x}+1}\right)^x=\left(\frac{1}{e}-1\right)^{\infty} \\
& =\text { (some negative value) }{ }^{\infty} \text { which is not defined as base is } \\
& \text {-ve. }
\end{aligned}
\)
\(\lim _{x \rightarrow 1} \frac{1-x^2}{\sin 2 \pi x}\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{1-x^2}{\sin 2 \pi x} \\
& =-\lim _{x \rightarrow 1} \frac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)} \\
& =-\lim _{x \rightarrow 1} \frac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \frac{1+x}{2 \pi}=\frac{-1}{\pi}
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{1}{x} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) is equal to
(d)
\(
\begin{aligned}
&\text { We know that } \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)= \begin{cases}2 \tan ^{-1} x, & x \geq 0 \\ -2 \tan ^{-1} x, & x \leq 0\end{cases}\\
&\begin{aligned}
& \lim _{x \rightarrow 0^{+}} \frac{1}{x} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\lim _{x \rightarrow 0^{+}} \frac{2 \tan ^{-1} x}{x}=2, \text { and } \\
& \lim _{x \rightarrow 0^{-}} \frac{1}{x} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\lim _{x \rightarrow 0^{+}}\left[-\frac{2 \tan ^{-1} x}{x}\right]=-2
\end{aligned}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty}\left(\frac{x^2+2 x-1}{2 x^2-3 x-2}\right)^{\frac{2 x+1}{2 x-1}}\) is equal to
(c)
\(
\lim _{x \rightarrow \infty}\left(\frac{x^2+2 x-1}{2 x^2-3 x-2}\right)^{\frac{2 x+1}{2 x-1}}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow \infty}\left(\frac{1+\frac{2}{x}-\frac{1}{x^2}}{2-\frac{3}{x}-\frac{2}{x^2}}\right)^{\frac{2+1 / x}{2-1 / x}} \\
& =1 / 2
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{2 \sqrt{x}+3 \sqrt[3]{x}+4 \sqrt[4]{x}+\cdots+n \sqrt[n]{x}}{\sqrt{(2 x-3)}+\sqrt[3]{(2 x-3)}+\cdots+\sqrt[n]{(2 x-3)}}\) is equal to
(c) Since the highest degree of \(x\) is \(1 / 2\), divide numerator and denominator by \(\sqrt{x}\), then we have limit \(\frac{2}{\sqrt{2}}\) or \(\sqrt{2}\).
\(\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}\) is equal to
(a)
\(
\begin{aligned}
& \lim _{y \rightarrow 0}\left\{\frac{x\{\sec (x+y)-\sec x\}}{y}+\sec (x+y)\right\} \\
& \lim _{y \rightarrow 0}\left[\frac{x}{y}\left\{\frac{\cos x-\cos (x+y)}{\cos (x+y) \cos x}\right\}\right]+\lim _{y \rightarrow 0} \sec (x+y) \\
& \lim _{y \rightarrow 0}\left[\frac{x 2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{y}{2}\right)}{y \cos (x+y) \cos x}\right]+\sec x \\
& \lim _{y \rightarrow 0}\left[\frac{x \sin \left(x+\frac{y}{2}\right)}{\cos (x+y) \cos x} \times \frac{\sin \left(\frac{y}{2}\right)}{\frac{y}{2}}\right]+\sec x \\
& \begin{array}{l}
x \tan x \sec x+\sec x \\
\sec x(x \tan x+1)
\end{array}
\end{aligned}
\)
The value of \(\lim _{m \rightarrow \infty}\left(\cos \frac{x}{m}\right)^m\) is
(a)
\(
\begin{aligned}
& \lim _{m \rightarrow \infty}\left(\cos \frac{x}{m}\right)^m \\
& =\lim _{m \rightarrow \infty}\left[1-\left(1-\cos \frac{x}{m}\right)\right]^m \\
& =\lim _{m \rightarrow \infty}\left[1-2 \sin ^2 \frac{x}{2 m}\right]^m \\
& =e^{\lim _{m \rightarrow \infty}\left(-2 \sin ^2 \frac{x}{2 m}\right)^m}=1
\end{aligned}
\)
\(\lim _{x \rightarrow 1}\left[\operatorname{cosec} \frac{\pi x}{2}\right]^{1 /(1-x)}\) (where [.] represents the greatest integer function) is equal to
(b)
\(
\begin{aligned}
& \operatorname{cosec} \frac{\pi x}{2} \rightarrow 1 \text { when } x \rightarrow 1 \Rightarrow\left[\operatorname{cosec} \frac{\pi x}{2}\right]=1 \\
& \therefore \quad \text { limit }=1
\end{aligned}
\)
\(\lim _{n \rightarrow \infty}\left(\frac{n^2-n+1}{n^2-n-1}\right)^{n(n-1)}\) is equal to
(b)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n^2-n+1}{n^2-n-1}\right)^{n(n-1)} \\
& =\lim _{n \rightarrow \infty}\left(\frac{n(n-1)+1}{n(n-1)-1}\right)^{n(n-1)}
\end{aligned}\\
&=\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n(n-1)}\right)^{n(n-1)}}{\left(1-\frac{1}{n(n-1)}\right)^{n(n-1)}}=\frac{e}{e^{-1}}=e^2
\end{aligned}
\)
If \(f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)\), then for \(x>0, y>0, f(x y)\) is equal to
\(
\begin{aligned}
& \text { (b) } f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right) \\
& =\lim _{n \rightarrow \infty} \frac{x^{1 / n}-1}{1 / n} \\
& =\lim _{m \rightarrow 0} \frac{x^m-1}{m}\left(\text { where } \frac{1}{n} \text { replaced by } m\right) \\
& =\ln x \\
& \Rightarrow f(x y)=\ln (x y)=\ln x+\ln y=f(x)+f(y)
\end{aligned}
\)
If \(\lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)\) exists, then
(c) If \(f(x)=\sin \left(\frac{1}{x}\right)\) and \(g(x)=\frac{1}{x}\), then both \(\lim _{x \rightarrow 0} f(x)\) and \(\lim _{x \rightarrow 0} g(x)\) do not exist, but \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=0\) exists.
If \(\lim _{n \rightarrow \infty} \frac{n \cdot 3^n}{n(x-2)^n+n \cdot 3^{n+1}-3^n}=\frac{1}{3}\), then the range of \(x\) is (where \(n \in N\) )
(a) \(\lim _{n \rightarrow \infty} \frac{n \cdot 3^n}{n(x-2)^n+n \cdot 3^{n+1}-3^n}=\frac{1}{3}\)
\(\lim _{n \rightarrow \infty} \frac{1}{\frac{(x-2)^n}{3^n \cdot}+3-\frac{1}{n}}\left(\right.\) Dividing \(N^r\) and \(D^r\) by \(\left.n \times 3^n\right)\)
For \(\lim _{n \rightarrow \infty}\) to be equal to \(1 / 3\)
\(\lim _{n \rightarrow \infty} \frac{1}{n} \rightarrow 0\) (which is true) and \(\lim _{n \rightarrow \infty}\left(\frac{x-2}{3}\right)^n \rightarrow 0\)
\(2 \leq x<5\)
\(
2 \leq x<5
\)
The value of \(\lim _{x \rightarrow 2} \frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}\) is
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}} \\
& =\lim _{x \rightarrow 2} \frac{\left(2^x\right)^2-6 \times 2^x+2^3}{\sqrt{2^x}-2} \quad\left[\text { Multiplying } N^{\prime} \text { and } D^{\prime} \text { by } 2^x\right] \\
& =\lim _{x \rightarrow 2} \frac{\left(2^x-4\right)\left(2^x-2\right)\left(\sqrt{2^x}+2\right)}{\left(\sqrt{2^x}-2\right)\left(\sqrt{2^x}+2\right)} \\
& =\lim _{x \rightarrow 2} \frac{\left(2^x-4\right)\left(2^x-2\right)\left(\sqrt{2^x}+2\right)}{\left(2^x-4\right)} \\
& =\lim _{x \rightarrow 2}\left(2^x-2\right)\left(\sqrt{2^x}+2\right)=\left(2^2-2\right)(2+2)=8
\end{aligned}
\)
\(\lim _{n \rightarrow \infty}\left(\left(\frac{n}{n+1}\right)^\alpha+\sin \frac{1}{n}\right)^n \quad(\) when \(\alpha \in Q)\) is equal to
(c) \(1^{\infty}\) form
\(
L=e^{\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^\alpha+\sin \frac{1}{n}-1\right)}=e^{\lim _{n \rightarrow \infty} n \sin \frac{1}{n}+\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^\alpha-1\right)}
\)
Consider, \(\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^\alpha-1\right)=\lim _{n \rightarrow \infty} n\left(\left(\frac{1}{1+1 / n}\right)^\alpha-1\right)\)
Put \(n=\frac{1}{y}\)
\(
\begin{aligned}
&=\lim _{y \rightarrow 0} \frac{1}{y}\left(\left(\frac{1}{1+y}\right)^\alpha-1\right)=\lim _{y \rightarrow 0} \frac{1-(1+y)^\alpha}{y}=-a\\
&\text { (Using binomial) }\\
&\therefore L=e^{1-\alpha}
\end{aligned}
\)
\(f(x)=\frac{\ln \left(x^2+e^x\right)}{\ln \left(x^4+e^{2 x}\right)}\). Then \(\lim _{n \rightarrow \infty} f(x)\) is equal to
(b)
\(
L=\lim _{x \rightarrow \infty} \frac{\ln \left(x^2+e^x\right)}{\ln \left(x^4+e^{2 x}\right)}=\lim _{x \rightarrow \infty} \frac{\ln e^x\left(1+\frac{x^2}{e^x}\right)}{\ln e^{2 x}\left(1+\frac{x^4}{e^{2 x}}\right)}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow \infty} \frac{x+\ln \left(1+\frac{x^2}{e^x}\right)}{2 x+\ln \left(1+\frac{x^4}{e^{2 x}}\right)} \\
& =\lim _{x \rightarrow \infty} \frac{1+\frac{1}{x} \ln \left(1+\frac{x^2}{e^x}\right)}{2+\frac{1}{x} \ln \left(1+\frac{x^4}{e^{2 x}}\right)}
\end{aligned}
\)
Note that \(\underset{x \rightarrow \infty}{\operatorname{as}} \frac{x^2}{e^x} \rightarrow 0[latex] and [latex]\underset{x \rightarrow \infty}{\operatorname{as}} \frac{x^2}{e^{2 x}} \rightarrow 0\) (Using L’Hopital’s rule)
Hence \(L=\frac{1}{2}\)
\(\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3 x}{1+x^2}\right)}{1+\cos \pi x}\) is equal to
(a)
\(
\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3 x}{1+x^2}\right)}{1+\cos \pi x}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{1-\cos \left(\frac{3 \pi}{2}-\frac{3 \pi x}{1+x^2}\right)}{1-\cos (\pi-\pi x)} \\
& =\lim _{x \rightarrow 1} \frac{2 \sin ^2\left(\frac{3 \pi}{4}-\frac{3 \pi x}{2\left(1+x^2\right)}\right)}{2 \sin ^2\left(\frac{\pi}{2}-\frac{\pi x}{2}\right)} \\
& =\lim _{x \rightarrow 1}\left(\frac{\frac{3 \pi}{4}-\frac{3 \pi x}{2\left(1+x^2\right)}}{\frac{\pi}{2}-\frac{\pi x}{2}}\right)^2 \\
& =\lim _{x \rightarrow 1} 9\left(\frac{\frac{1}{2}-\frac{x}{1+x^2}}{1-x}\right)^2=\lim _{x \rightarrow 1} 9\left(\frac{x-1}{2\left(1+x^2\right)}\right)^2=0
\end{aligned}
\)
\(\lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10)\) is equal to
(b)
\(
\because \lim _{n \rightarrow \infty} \cos ^{2 n} x=\left\{\begin{array}{l}
1, x=r \pi, r \in I \\
0, x \neq r \pi, r \in I
\end{array}\right.
\)
Here, for \(x=10, \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1\) and in all other cases it is zero.
\(
\therefore \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1
\)
The value of \(\lim _{x \rightarrow \infty} \frac{\left(2^{x^n}\right)^{\frac{1}{e^x}}-\left(3^{x^n}\right)^{\frac{1}{e^x}}}{x^n}(\) where \(n \in N)\) is
(b)
\(
L=\lim _{x \rightarrow \infty} \frac{\left(2^{x^n}\right)^{\frac{1}{x}}-\left(3^{x^n}\right)^{\frac{1}{x}}}{x^n}=\lim _{x \rightarrow \infty} \frac{(3)^{\frac{x^n}{x}}\left(\left(\frac{2}{3}\right)^{\frac{x^n}{e^x}}-1\right)}{x^n}
\)
Now, \(\lim _{x \rightarrow \infty} \frac{x^n}{e^x}=\lim _{x \rightarrow \infty} \frac{n!}{e^x}=0 \quad\) (differentiating numerator and denominator \(n\) times for L’Hopital’s rule)
\(
\begin{aligned}
& \text { Hence } L=\lim _{x \rightarrow \infty}(3)^{\frac{x^n}{x^x}} \lim _{x \rightarrow \infty} \frac{\left(\left(\frac{2}{3}\right)^{\frac{x^n}{e^x}}-1\right)}{\frac{x^n}{e^x}} \lim _{x \rightarrow \infty} \frac{1}{e^x} \\
& =1 \times \log (2 / 3) \times 0=0
\end{aligned}
\)
If \(f:(1,2) \rightarrow R\) satisfies the inequality \(\frac{\cos (2 x-4)-33}{2}<f(x)<\frac{x^2|4 x-8|}{x-2}, \forall x \in(1,2)\), then \(\lim _{x \rightarrow 2^{-}} f(x)\) is
(c)
\(
\begin{aligned}
& \frac{\cos (2 x-4)-33}{2}<f(x)<\frac{x^2|4 x-8|}{x-2} \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{\cos (2 x-4)-33}{2}<\lim _{x \rightarrow 2^{-}} f(x)<\lim _{x \rightarrow 2^{-}} \frac{x^2|4 x-8|}{x-2} \\
& \Rightarrow-16<\lim _{x \rightarrow 2^{-}} f(x)<\lim _{x \rightarrow 2^{-}} \frac{x^2(8-4 x)}{x-2} \\
& \Rightarrow-16<\lim _{x \rightarrow 2^{-}} f(x)<-16 \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} f(x)=-16 \text { (by sandwich theorem) }
\end{aligned}
\)
Let \(f(x)=\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5}\). Then the set of values of \(x\) for which \(f(x)=0\) is
(a)
\(
\begin{aligned}
& \text { Given } g(x)=\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5}=0 \\
& \Rightarrow\left[\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^2\right]^n \rightarrow \infty \\
& \Rightarrow\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^2>1 \\
& \Rightarrow\left|\tan ^{-1} 2 x\right|>\frac{\pi}{3} \\
& \Rightarrow \tan ^{-1} 2 x<-\frac{\pi}{3} \text { or } \tan ^{-1} 2 x>\frac{\pi}{3} \\
& \Rightarrow 2 x<-\sqrt{3} \text { or } 2 x>\sqrt{3} \Rightarrow|2 x|>\sqrt{3}
\end{aligned}
\)
\(\lim _{x \rightarrow 0}\left\{(1+x)^{\frac{2}{x}}\right\}\) (where \(\{\cdot\}\) denotes the fractional part of \(x\) ) is equal to
(a)
\(
\begin{aligned}
&(1+x)^{2 / x}=(1+x)^{2 / x}-\left[(1+x)^{2 / x}\right]\\
&\text { Now, } \lim _{x \rightarrow 0}(1+x)^{2 / x}=e^2\\
&\Rightarrow \lim _{x \rightarrow 0}\left\{(1+x)^{2 / x}\right\}=e^2-\left[e^2\right]=e^2-7
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{\sin \left(x^2\right)}{\ln \left(\cos \left(2 x^2-x\right)\right)}\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin \left(x^2\right)}{\ln \left(\cos \left(2 x^2-x\right)\right)} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(x^2\right)}{\log \left(1-2 \sin ^2\left(\frac{2 x^2-x}{2}\right)\right)} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(x^2\right) x^2}{x^2 \log \left(1-2 \sin ^2\left(\frac{2 x^2-x}{2}\right)\right)}\left[-2 \sin ^2\left(\frac{2 x^2-x}{2}\right)\right] \\
& =\lim _{x \rightarrow 0}-\frac{-2 \sin ^2\left(\frac{2 x^2-x}{2}\right)}{\frac{2 \sin ^2\left(\frac{2 x^2-x}{2}\right)}{\left(\frac{2 x^2-x}{2}\right)^2}\left(\frac{2 x^2-x}{2}\right)^2} \\
& =\lim _{x \rightarrow 0}-\frac{2 x^2}{\left(2 x^2-x\right)^2}=\lim _{x \rightarrow 0}-\frac{2}{(2 x-1)^2}=-2
\end{aligned}
\)
\(\lim _{x \rightarrow-1} \frac{1}{\sqrt{|x|-\{-x\}}}\) (where \(\{x\}\) denotes the fractional part of \(x\) ) is equal to
(b)
\(
\begin{aligned}
&\begin{aligned}
& \text { L.H.L. }=\lim _{x \rightarrow-1^{-}} \frac{1}{\sqrt{|x|-\{-x\}}}=\lim _{x \rightarrow-1^{-}} \frac{1}{\sqrt{-x-(x+2)}} \\
& =\lim _{x \rightarrow-1^{-}} \frac{1}{\sqrt{-2 x-2}}=\infty \\
& \text { R.H.L. }=\lim _{x \rightarrow-1^{+}} \frac{1}{\sqrt{|x|-\{-x\}}}=\lim _{x \rightarrow-1^{-}} \frac{1}{\sqrt{-x-(x+1)}} \\
& =\lim _{x \rightarrow-1^{-}} \frac{1}{\sqrt{-2 x-1}}=1
\end{aligned}\\
&\text { Hence, the limit does not exist. }
\end{aligned}
\)
If \(f(x)=\left\{\begin{array}{ll}x^n \sin \left(1 / x^2\right), & x \neq 0 \\ 0, & x=0\end{array},(n \in I)\right.\), then
(a)
\(
\begin{aligned}
&\text { For } n>1 \text {, }\\
&\begin{aligned}
& \lim _{x \rightarrow 0} x^n \sin \left(1 / x^2\right)=0 \times(\text { any value between }-1 \text { to } 1)=0 \\
& \text { For } n<0 \\
& \lim _{x \rightarrow 0} x^n \sin \left(1 / x^2\right)=\infty \times(\text { any value between }-1 \text { to } 1)=\infty
\end{aligned}
\end{aligned}
\)
The value of \(\lim _{x \rightarrow 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right) ; p, q, \in N\) equals
(c)
\(
\lim _{x \rightarrow 1} \frac{p-q+q x^p-p x^q}{1-x^p-x^q+x^{p+q}}\left(\frac{0}{0}\right)
\)
\(
=\lim _{x \rightarrow 1} \frac{p q x^{p-1}-p q x^{q-1}}{-p x^{p-1}-q x^{q-1}+(p+q) x^{p+q-1}}\left(\frac{0}{0}\right)(\text { L’ Hopital Rule })
\)
\(
\begin{aligned}
&=\lim _{x \rightarrow 1} \frac{p q(p-1) x^{p-2}-p q(q-1) x^{q-2}}{-p(p-1) x^{p-2}-q(q-1) x^{q-2}+(p+q)(p+q-1) x^{p+q-2}}\\
&\text { (L’ Hopital rule) }
\end{aligned}
\)
\(
=\frac{p-q}{2}
\)
latex]\lim _{x \rightarrow-1}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\frac{1-\cos (x+1)}{(x+1)^2}}[/latex] is equal to:
(b)
\(
\lim _{x \rightarrow-1}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\frac{1-\cos (x+1)}{(x+1)^2}}
\)
\(
=\lim _{x \rightarrow-1}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\lim _{x \rightarrow-1} \frac{1-\cos (x+1)}{(x+1)^2}}
\)
\(
=\left(\frac{2}{3}\right)^{\lim _{x \rightarrow-1} \frac{\sin (x+1)}{2(x+1)}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}
\)
The value of \(\lim _{x \rightarrow 2}\left(\left(\frac{x^3-4 x}{x^3-8}\right)^{-1}-\left(\frac{x+\sqrt{2 x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right)\) is
(a)
\(
\begin{aligned}
& \lim _{x \rightarrow 2}\left[\left(\frac{x^3-4 x}{x^3-8}\right)^{-1}\right. \\
& \left.\quad-\left(\frac{\sqrt{x}(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right] \\
& =\lim _{x \rightarrow 2}\left[\frac{x^2+2 x+4}{x(x+2)}-\left(\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right] \\
& =\lim _{x \rightarrow 2}\left[\frac{x^2+2 x+4}{x(x+2)}-1\right]=\frac{12}{8}-1=\frac{1}{2}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{e^{1 / x^2}-1}{2 \tan ^{-1}\left(x^2\right)-\pi}\) is equal to
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{e^{1 / x^2}-1}{2 \tan ^{-1}\left(x^2\right)-\pi} \\
& =\lim _{t \rightarrow 0^{+}} \frac{e^{t^2}-1}{2 \cot ^{-1} t^2-\pi} \\
& =\lim _{t \rightarrow 0^{+}} \frac{e^{t^2}-1}{-2 \tan t^2} \\
& =\lim _{t \rightarrow 0^{+}}-\frac{1}{2} \frac{e^{t^2}-1}{t^2 \frac{\tan t^2}{t^2}}=-\frac{1}{2}
\end{aligned}
\)
The value of \(\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x+\log (1-x)}{x^3}\) is
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1+\sin x-\cos x+\log (1-x)}{x^3} \\
& =\lim _{x \rightarrow 0} \frac{1+\left(x-\frac{x^3}{3!}+\cdots\right)-\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots\right)}{x^3} \\
& =-\frac{1}{3!}-\frac{1}{3}=-\frac{1}{2}
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{\cos (\tan x)-\cos x}{x^4}\) is equal to
\(
\begin{aligned}
& \text { (b) } \cos (\tan x)-\cos x=2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right) \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\cos (\tan x)-\cos x}{x^4}=\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)}{x^4} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)}{x^4\left(\frac{x+\tan x}{2}\right)\left(\frac{x-\tan x}{2}\right)}\left(\frac{x^2-\tan ^2 x}{4}\right) \\
& =\frac{1}{2} \lim _{x \rightarrow 0} \frac{x^2-\tan ^2 x}{x^4} \\
& =\frac{1}{2} \lim _{x \rightarrow 0} \frac{x^2-\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\cdots\right)^2}{x^4} \\
& =\frac{1}{2} \lim _{x \rightarrow 0} \frac{1}{x^2}\left(1-\left(1+\frac{x^2}{3}+\frac{2}{15} x^4+\cdots\right)^2\right)=-\frac{1}{3}
\end{aligned}
\)
If \(x_1=3\) and \(x_{n+1}=\sqrt{2+x_n}, n \geq 1\), then \(\lim _{n \rightarrow \infty} x_n\) is
(b)
\(
\begin{aligned}
& x_{n+1}=\sqrt{2+x_n} \\
& \lim _{n \rightarrow \infty} x_{n+1}=\sqrt{2+\lim _{n \rightarrow \infty} x_n}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow t=\sqrt{2+t} \left(\because \lim _{x \rightarrow \infty} x_{n+1}=\lim _{x \rightarrow \infty} x_n=t\right) \\
& \Rightarrow t^2-t-2=0 \\
& \Rightarrow(t-2)(t+1)=0 \\
& \Rightarrow t=2 \left(\because x_n>0 \forall n \therefore t>0\right)
\end{aligned}
\)
\(\lim _{x \rightarrow 0}\left(\frac{1^x+2^x+3^x+\cdots+n^x}{n}\right)^{1 / x}\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{1^x+2^x+\cdots+n^x}{n}\right)^{1 / x} \\
& =e^{\lim _{x \rightarrow 0}\left(\frac{1^x-1}{n}+\frac{2^x-1}{n}+\cdots+\frac{n^x-1}{n}\right) \frac{1}{x}} \\
& =e^{\lim _{x \rightarrow 0} \frac{1}{n}\left\{\frac{1^x-1}{x}+\frac{2^x-1}{x}+\cdots+\frac{n^x-1}{x}\right\}} \\
& =e^{\frac{1}{n}[\log 1+\log 2+\cdots+\log n]} \\
& =e^{\frac{1}{n}(\log n!)}=e^{\log (n!)^{\frac{1}{n}}}=(n!)^{\frac{1}{n}}
\end{aligned}
\)
The value of the limit \(\lim _{x \rightarrow 0} \frac{a^{\sqrt{x}}-a^{1 / \sqrt{x}}}{a^{\sqrt{x}}+a^{1 / \sqrt{x}}}, a>1\) is
\(
\begin{aligned}
\text { (c) } & \lim _{x \rightarrow 0} \frac{a^{\sqrt{x}}-a^{1 / \sqrt{x}}}{a^{\sqrt{x}}+a^{1 / \sqrt{x}}}, a>1 \\
& \text { Put } x=t^2 \\
\therefore & \lim _{t \rightarrow 0} \frac{a^t-a^{1 / t}}{a^t+a^{1 / t}} \\
\Rightarrow & \lim _{t \rightarrow 0} \frac{a^{t-1 / t}-1}{a^{t-1 / t}+1}=\frac{a^{-\infty}-1}{a^{-\infty}+1}=\frac{0-1}{0+1}=-1
\end{aligned}
\)
Among (i) \(\lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{x}{\sin x}\right)\) and
(ii) \(\lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{\sin x}{x}\right)\)
(a)
(i)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{x}{\sin x}\right) \\
& =\sec ^{-1}\left(\frac{\infty}{\sin \infty}\right) \\
& =\sec ^{-1}\left(\frac{\infty}{\text { any value between }-1 \text { to } 1}\right) \\
& =\sec ^{-1}( \pm \infty)=\frac{\pi}{2}
\end{aligned}
\)
(ii)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{\sin x}{x}\right)=\sec ^{-1}\left(\frac{\sin \infty}{\infty}\right) \\
& \sec ^{-1}\left(\frac{\text { any value between }-1 \text { to } 1}{\infty}\right) \\
& \sec ^{-1} 0=\text { not defined }
\end{aligned}
\)
Hence (i) exists but (ii) does not exist.
If \(\lim _{x \rightarrow 0} \frac{x^n-\sin x^n}{x-\sin ^n x}\) is non-zero finite, then \(n\) must be equal
(b) For \(n=0\), we have \(\lim _{x \rightarrow 0} \frac{1-\sin 1}{x-1}=\sin 1-1\)
For \(n=1, \lim _{x \rightarrow 0} \frac{x-\sin x}{x-\sin x}=1\)
For \(n=2, \lim _{x \rightarrow 0} \frac{x^2-\sin ^2 x}{x-\sin ^2 x}=\lim _{x \rightarrow 0} \frac{1-\frac{\sin ^2 x}{x^2}}{\frac{1}{x}-\frac{\sin ^2 x}{x^2}}\)
This does not exist.
For \(n=3\) also given limit does not exist.
Hence \(n=0\) or 1.
If \(\lim _{x \rightarrow-2^{-}} \frac{a e^{1 /|x+2|}-1}{2-e^{1 /|x+2|}}=\lim _{x \rightarrow-2^{+}} \sin \left(\frac{x^4-16}{x^5+32}\right)\), then \(a\) is.
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow-2^{-}} \frac{a e^{1 /|x+2|}-1}{2-e^{1 /|x+2|}}=\lim _{x \rightarrow-2^{-}} \frac{a-e^{-1 /|x+2|}}{2 e^{-1 /|x+2|}-1}=-a \\
& \lim _{x \rightarrow-2^{-}} \sin \left(\frac{x^4-16}{x^5+32}\right)=\lim _{x \rightarrow-2^{-}} \sin \left(\frac{\frac{x^4-(-2)^4}{x-(-2)}}{\frac{x^5-(-2)^5}{x-(-2)}}\right) \\
& =\sin \left(-\frac{2}{5}\right) \Rightarrow a=\sin \frac{2}{5}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty}\left((x+5) \tan ^{-1}(x+5)-(x+1) \tan ^{-1}(x+1)\right)\) is equal to
\(
\begin{aligned}
& \text { (b) Given limit is } \lim _{x \rightarrow \infty}(x+1)\left[\tan ^{-1}(x+5)-(x+1)\right]+ \\
& =\lim _{x \rightarrow \infty}\left[(x+1) \tan ^{-1} \frac{4}{1+(x+1)(x+5)}+4 \tan ^{-1}(x+5)\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{x \rightarrow \infty}\left[(x+1) \tan ^{-1} \frac{\frac{4}{x^2+6 x+6}}{\left(\frac{4}{x^2+6 x+6}\right)} \times \frac{4}{x^2+6 x+6}+4 \tan ^{-1}(x+5)\right] \\
& =0+4 \times \frac{\pi}{2}=2 \pi
\end{aligned}
\)
\(
\lim _{x \rightarrow 1} \frac{(1-x)\left(1-x^2\right) \cdots\left(1-x^{2 n}\right)}{\left\{(1-x)\left(1-x^2\right) \cdots\left(1-x^n\right)\right\}^2}, n \varepsilon N
\)
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{(1-x)\left(1-x^2\right) \cdots\left(1-x^{2 n}\right)}{\left\{(1-x)\left(1-x^2\right) \cdots\left(1-x^n\right)\right\}^2} \\
& =\lim _{x \rightarrow 1} \frac{\left(\frac{1-x}{1-x}\right)\left(\frac{1-x^2}{1-x}\right) \cdots\left(\frac{1-x^{2 n}}{1-x}\right)}{\left(\left(\frac{1-x}{1-x}\right)\left(\frac{1-x^2}{1-x}\right) \cdots\left(\frac{1-x^n}{1-x}\right)\right)^2} \\
& =\frac{1 \times 2 \times 3 \cdots(2 n)}{(1 \times 2 \times 3 \cdots n)^2}=\frac{(2 n)!}{n!n!}={ }^{2 n} \mathrm{C}_n
\end{aligned}
\)
The value of \(\lim _{x \rightarrow 0}\left(\left[\frac{100 x}{\sin x}\right]+\left[\frac{99 \sin x}{x}\right]\right)\) (where [.] represents the greatest integral function) is
The value of \(\lim _{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)}\) is
\(
\begin{aligned}
&\text { (a) Let } \sin ^{-1} x=\theta \text {. Then, } x=\sin \theta \text {. }\\
&\begin{aligned}
& \text { Now, } x \rightarrow \frac{1}{\sqrt{2}} \Rightarrow \sin \theta \rightarrow \frac{1}{\sqrt{2}} \Rightarrow \theta \rightarrow \frac{\pi}{4} \\
\therefore & \lim _{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)} \\
= & \lim _{\theta \rightarrow \frac{\pi}{4}} \frac{\sin \theta-\cos \theta}{1-\tan \theta} \\
= & \lim _{\theta \rightarrow \frac{\pi}{4}} \frac{(\sin \theta-\cos \theta)}{(\cos \theta-\sin \theta)} \cos \theta
\end{aligned}
\end{aligned}
\)
\(
=\lim _{\theta \rightarrow \frac{\pi}{4}}-\cos \theta=-\frac{1}{\sqrt{2}}
\)
The value of \(\lim _{x \rightarrow 1^{-}} \frac{1-\sqrt{x}}{\left(\cos ^{-1} x\right)^2}\) is
\(
\begin{aligned}
&\text { (d) We have } \lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{\left(\cos ^{-1} x\right)^2}\\
&\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{(1-\sqrt{x})(1+\sqrt{x})}{\left(\cos ^{-1} x\right)^2(1+\sqrt{x})} \\
& =\lim _{x \rightarrow 1} \frac{1-x}{\left(\cos ^{-1} x\right)^2(1+\sqrt{x})} \\
& =\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2(1+\sqrt{\cos \theta})} 1, \text { where } x=\cos \theta \\
& =\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} \frac{1}{(1+\sqrt{\cos \theta})} \\
& \left.=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 \frac{\theta}{2}}{4 \frac{\theta^2}{4}}\left(\frac{1}{1+\sqrt{\cos \theta}}\right) \Rightarrow \cos \theta \rightarrow 1 \Rightarrow \theta \rightarrow 0\right] \\
& =\frac{1}{2} \lim _{\theta \rightarrow 0}\left(\frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}\right)^2 \frac{1}{(1+\sqrt{\cos \theta})}=\frac{1}{2}(1)^2 \frac{1}{(1+1)}=\frac{1}{4}
\end{aligned}
\end{aligned}
\)
\(\lim _{x \rightarrow 0}\left[\min \left(y^2-4 y+11\right) \frac{\sin x}{x}\right]\) (where [.] denotes the greatest integer function) is
\(
\begin{aligned}
& \text { (b) } \min \left(y^2-4 y+11\right)=\min \left[(y-2)^2+7\right]=7 \\
& \Rightarrow L=\lim _{x \rightarrow 0}\left[\min \left(y^2-4 y+11\right) \frac{\sin x}{x}\right] \\
& =\lim _{x \rightarrow 0}\left[\frac{7 \sin x}{x}\right] \\
& =[\text { a value slightly lesser than } 7] \quad(|\sin x|<|x|, \text { when } x \rightarrow 0) \\
& \Rightarrow L=\lim _{x \rightarrow 0}\left[7 \frac{\sin x}{x}\right]=6
\end{aligned}
\)
\(\lim _{x \rightarrow \pi / 2} \frac{\sin (x \cos x)}{\cos (x \sin x)}\) is equal to
\(
\begin{aligned}
& \text { (b) } L=\lim _{x \rightarrow \pi / 2} \frac{\sin (x \cos x)}{\sin \left(\frac{\pi}{2}-x \sin x\right)} \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (x \cos x)}{(x \cos x)} \frac{x \cos x}{\sin \left(\frac{\pi}{2}-x \sin x\right)} \frac{\left(\frac{\pi}{2}-x \sin x\right)}{\left(\frac{\pi}{2}-x \sin x\right)} \\
& =1 \times 1 \lim _{x \rightarrow \pi / 2} \frac{x \cos x}{\left(\frac{\pi}{2}-x \sin x\right)}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Put } x=\pi / 2+h\\
&\begin{aligned}
& \text { Then, } L=\lim _{h \rightarrow 0} \frac{\left(\frac{\pi}{2}+h\right) \cos \left(\frac{\pi}{2}+h\right)}{\frac{\pi}{2}-\left(\frac{\pi}{2}+h\right) \sin \left(\frac{\pi}{2}+h\right)} \\
& =\lim _{h \rightarrow 0} \frac{-\left(\frac{\pi}{2}+h\right) \sin h}{\frac{\pi}{2}(1-\cos h)-h \cos h}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{-\left(\frac{\pi}{2}+h\right)\left(\frac{\sin h}{h}\right)}{\frac{\pi}{2} \frac{(1-\cos h)}{h}-\cos h}\\
&\text { (Divide } N^r \text { and } D^r \text { by } h \text { ) }
\end{aligned}
\)
\(
=\frac{-\left(\frac{\pi}{2}+0\right) 1}{0-1}=\frac{\pi}{2}
\)
If \(\lim _{x \rightarrow 0}\left(x^{-3} \sin 3 x+a x^{-2}+b\right)\) exists and is equal to 0, then
\(
\begin{aligned}
& \text { (a) } \lim _{x \rightarrow 0} \frac{\sin 3 x}{x^3}+\frac{a}{x^2}+b \\
& =\lim _{x \rightarrow 0} \frac{\sin 3 x+a x+b x^3}{x^3} \\
& =\lim _{x \rightarrow 0} \frac{3 \frac{\sin 3 x}{3 x}+a+b x^2}{x^2} \\
& \quad \text { For existence, }(3+a)=0 \\
& \Rightarrow a=-3 \\
& \therefore L=\lim _{x \rightarrow 0} \frac{\sin 3 x-3 x+b x^3}{x^3} \\
& =27 \lim _{t \rightarrow 0} \frac{\sin t-t}{t^3}+\mathrm{b}=0(3 x=\mathrm{t}) \\
& =-\frac{27}{6}+b=0 \\
& \Rightarrow b=\frac{9}{2}
\end{aligned}
\)
If \(\lim _{x \rightarrow 0} \frac{x^n \sin ^n x}{x^n-\sin ^n x}\) is non-zero finite, then \(n\) is equal to
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x^n \sin ^n x}{x^n-\sin ^n x} \\
& \lim _{x \rightarrow 0} \frac{x^n\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)^n}{x^n-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)^n} \\
& \lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)^n}{1-\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^n} \\
& =\lim _{x \rightarrow 0} \frac{x^n\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^n}{1-\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^n}
\end{aligned}
\)
\(
\begin{aligned}
&\text { For } n=2 \text {, }\\
&\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x^2\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^2}{1-\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^2} \\
\Rightarrow & \lim _{x \rightarrow 0} \frac{x^2\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)^2}{\left(2-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)\left(\frac{x^2}{3!}-\frac{x^4}{5!}+\cdots\right)} \\
= & \frac{1(1-0+\cdots)^2}{(2-0+0)\left(\frac{1}{3!}-0+\cdots\right)} \\
= & 3
\end{aligned}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{\left(1+x+x^2\right)}{x(\ln x)^3}\) is equal to
(d)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{1+x+x^2}{x(\ln x)^3}=\lim _{t \rightarrow 0^{+}} \frac{t^2+t+1}{t^2 \frac{1}{t}\left(\ln \left(\frac{1}{t}\right)\right)^3} \\
& \quad=\lim _{t \rightarrow 0^{+}} \frac{1+t+t^2}{-t(\ln t)^3}=+\infty
\end{aligned}
\)
\(\lim _{x \rightarrow 0} \frac{\left(2^m+x\right)^{1 / m}-\left(2^n+x\right)^{1 / n}}{x}\) is equal to
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(2^m+x\right)^{1 / m}-\left(2^n+x\right)^{1 / n}}{x} \\
& =\lim _{x \rightarrow 0} \frac{\left(2^m+x\right)^{1 / m}-2}{x}-\lim _{x \rightarrow 0} \frac{\left(2^n+x\right)^{1 / n}-2}{x} \\
& =\lim _{a \rightarrow 2} \frac{a-2}{a^m-2^m}-\lim _{b \rightarrow 2} \frac{b-2}{b^n-2^n} \text { [Putting } 2^m+x=a^m \text { and } 2^n+x=b^n \text { ] }
\end{aligned}
\)
\(
=\frac{1}{m 2^{m-1}}-\frac{1}{n 2^{n-1}}
\)
\(\lim _{x \rightarrow 0}\left[\left(1-e^x\right) \frac{\sin x}{|x|}\right]\) is (where [.] represents the greatest integer function)
(a)
\(
\begin{aligned}
& \lim _{x \rightarrow 0^{+}}\left[\left(1-e^x\right) \frac{\sin x}{|x|}\right] \\
& =\lim _{x \rightarrow 0^{+}}\left[\left(0^{-}\right) \frac{\sin x}{x}\right]=\left[0^{-}\right]=-1 \\
& =\lim _{x \rightarrow 0^{-}}\left[\left(1-e^x\right) \frac{\sin x}{|x|}\right] \\
& =\lim _{x \rightarrow 0^{-}}\left[\left(0^{+}\right) \frac{\sin x}{-x}\right]=\left[0^{-}\right]=-1 \\
& \text { Hence } \lim _{x \rightarrow 0}\left[\left(1-e^x\right) \frac{\sin x}{|x|}\right]=-1
\end{aligned}
\)
Let \(f(x)=\left\{\begin{array}{ll}x+1, & x>0 \\ 2-x, & x \leq 0\end{array}\right.\) and
\(
g(x)=\left\{\begin{array}{lc}
x+3, & x<1 \\
x^2-2 x-2, & 1 \leq x<2, \text { then } \lim _{x \rightarrow 0} g(f(x)) \text { is }\\
x-5, & x \geq 2
\end{array}\right.
\)
(c)
\(
\begin{aligned}
& \text { As } x \rightarrow 0^{-} \Rightarrow f(x) \rightarrow f\left(0^{-}\right)=2^{+} \\
& \lim _{x \rightarrow 0^{-}} g(f(x))=g\left(2^{+}\right)=-3
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Also as } x \rightarrow 0^{+} \Rightarrow f(x) \rightarrow f\left(0^{+}\right)=1^{+} \\
\Rightarrow & \lim _{x \rightarrow 0^{+}} g(f(x))=g\left(1^{+}\right)=-3
\end{aligned}\\
&\text { Hence } \lim _{x \rightarrow 0} g(f(x)) \text { exists and is equal to }-3\\
&\Rightarrow \lim _{x \rightarrow 0} g(f(x))=-3
\end{aligned}
\)
\(\lim _{x \rightarrow 1} \frac{n x^{n+1}-(n+1) x^n+1}{\left(e^x-e\right) \sin \pi x}\), where \(n=100\) is equal to
(c)
\(
\begin{aligned}
&I=\lim _{x \rightarrow 1} \frac{n x^n(x-1)-\left(x^n-1\right)}{\left(e^x-e\right) \sin \pi x}\\
&\text { put } x=1+h \text { so that as } x \rightarrow 1, h \rightarrow 0\\
&I=-\lim _{h \rightarrow 0} \frac{h \cdot n(1+h)^n-\left((1+h)^n-1\right)}{e\left(e^h-1\right) \sin \pi h}\\
&I=-\lim _{x \rightarrow 1} \frac{n \cdot h\left(1+{ }^n C_1 h+{ }^n C_2 h^2+{ }^n C_3 h^3+\cdots\right)}{\pi e\left(h^2\right)\left(\frac{e^h-1}{h}\right)}\frac{-\left(1+{ }^n C_1 h++{ }^n C_2 h^2++{ }^n C_3 h^3+\cdots-1\right)}{\left(\frac{\sin \pi h}{\pi h}\right)}
\end{aligned}
\)
\(
\begin{aligned}
= & -\frac{n^2-{ }^n C_2}{\pi e}=-\left[\frac{2 n^2-n(n-1)}{2 \pi e}\right]=-\frac{n^2+n}{2(\pi e)}=-\frac{n(n+1)}{2(\pi e)} \\
& \text { if } n=100 \Rightarrow 1=-\left(\frac{5050}{\pi e}\right)
\end{aligned}
\)
The value of \(\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{e^{1 / n}}{n}+\frac{e^{2 / n}}{n}+\ldots .+\frac{e^{(n-1) / n}}{n}\right]\) is
(c)
\(
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{e^{1 / n}}{n}+\frac{e^{2 / n}}{n}+\cdots+\frac{e^{(n-1) / n}}{n}\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{1+e^{1 / n}+\left(e^{1 / n}\right)^2+\cdots+\left(e^{1 / n}\right)^{n-1}}{n}\right] \\
& =\lim _{n \rightarrow \infty} \frac{1 \cdot\left[\left(e^{1 / n}\right)^n-1\right]}{n\left(e^{1 / n}-1\right)}=(e-1) \lim _{n \rightarrow \infty} \frac{1}{\left(\frac{e^{1 / n}-1}{1 / n}\right)} \\
& =(e-1) \times 1=(e-1)
\end{aligned}
\)
The value of \(\lim _{n \rightarrow \infty}\left[\frac{2 n}{2 n^2-1} \cos \frac{n+1}{2 n-1}-\frac{n}{1-2 n} \cdot \frac{n(-1)^n}{n^2+1}\right]\) is
(c)
\(
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{2}{2-\frac{1}{n^2}} \cdot \frac{1}{n} \cos \left(\frac{1+1 / n}{2-1 / n}\right)-\frac{1}{\left(\frac{1}{n}-2\right)} \cdot \frac{(-1)^n}{\left(1+\frac{1}{n^2}\right)^n} \cdot \frac{1}{n}\right] \\
& \quad=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{2}{2-\frac{1}{n^2}} \cdot \cos \left(\frac{1+\frac{1}{n}}{2-\frac{1}{n}}\right)-\frac{1}{\left(\frac{1}{n}-2\right)} \cdot \frac{(-1)^n}{\left(1+\frac{1}{n^2}\right)}\right] \\
& =0 \times\left[\frac{2}{2} \times \cos \frac{1}{2}+\frac{1}{2} \times \frac{1}{1}\right]=0 .
\end{aligned}
\)
\(
\lim _{x \rightarrow 0} \frac{\log \left(1+x+x^2\right)+\log \left(1-x+x^2\right)}{\sec x-\cos x}=
\)
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\log \left(1+x+x^2\right)+\log \left(1-x+x^2\right)}{\sec x-\cos x} \\
& =\lim _{x \rightarrow 0} \frac{\log \left[\left(1+x^2\right)^2-x^2\right]}{\left(1-\cos ^2 x\right) / \cos x} \\
& =\lim _{x \rightarrow 0} \frac{\log \left(1+x^2+x^4\right)}{\sin x \tan x} \\
& =\lim _{x \rightarrow 0} \frac{\log \left(1+x^2\left(1+x^2\right)\right)}{x^2\left(1+x^2\right)} \cdot x^2\left(1+x^2\right) \cdot \frac{1}{\frac{\sin x}{x} \cdot \frac{\tan x}{x} \cdot x^2} \\
& =1 \cdot\left(\operatorname{as} \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right)
\end{aligned}
\)
The value of \(\lim _{x \rightarrow a} \sqrt{a^2-x^2} \cot \frac{\pi}{2} \sqrt{\frac{a-x}{a+x}}\) is
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow a} \sqrt{a^2-x^2} \cdot \cot \frac{\pi}{2} \sqrt{\frac{a-x}{a+x}} \\
& =\lim _{x \rightarrow a} \frac{\sqrt{a^2-x^2}}{\tan \frac{\pi}{2} \sqrt{\frac{a-x}{a+x}}} \\
& =\frac{2}{\pi} \lim _{x \rightarrow a} \frac{\frac{\pi}{2} \sqrt{\frac{a-x}{a+x}}}{\tan \frac{\pi}{2} \sqrt{\frac{a-x}{a+x}}}(a+x)=\frac{4 a}{\pi}
\end{aligned}
\)
\(\lim _{x \rightarrow \infty} \frac{\cot ^{-1}\left(x^{-a} \log _a x\right)}{\sec ^{-1}\left(a^x \log _x a\right)}(a>1)\) is equal to
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\cot ^{-1}\left(x^{-a} \log _a x\right)}{\sec ^{-1}\left(a^x \log _x a\right)}(a>1) \\
& =\lim _{x \rightarrow \infty} \frac{\cot ^{-1}\left(\frac{\log _a x}{x^a}\right)}{\sec ^{-1}\left(\frac{a^x}{\log _a x}\right)} \operatorname{as}_{x \rightarrow \infty}\left(\frac{\log _a x}{x^a}\right) \rightarrow 0
\end{aligned}
\)
\(
\begin{aligned}
& \text { and }\left(\frac{a^x}{\log _a x}\right) \rightarrow \infty \text { (using L’Hopital rule) } \\
& \therefore I=\frac{\pi / 2}{\pi / 2}=1
\end{aligned}
\)
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