Chapter 5

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2}(\tan (\sin x))\right)}{x^{2}}\) is equal to (a) \(\pi\) (b) \(\frac{\pi}{4}\) (c) \(\frac{\pi}{2}\) (d) None of these

\(\lim _{t \rightarrow 0} \frac{1-(1+t)^{t}}{\ln (1+t)-t}\) is equal to (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 2 (d) \(-2\)

If \(I_{1}=\lim _{x \rightarrow \infty}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x\) and \(I_{2}=\lim _{x \rightarrow 0}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x\), then \(\left(I_{1}, I_{2}\right)\) is (a) \((0,0)\) (b) \((0,1)\) (c) \((1,0)\) (d) None of these

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 (a) 0 (b) \(\pi\) (c) \(2 \pi\) (d) None of these

\(\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\sin ^{2} x}-\sqrt[4]{1-2 \tan x}}{\sin x+\tan ^{2} x}\) is equal to (a) \(-1\) (b) 1 (c) \(\frac{1}{2}\) (d) \(-\frac{1}{2}\)

If \(x_{n+1}=\sqrt{\frac{1+x_{n}}{2}}\) and \(\left|x_{0}\right|<1, n \geq 0\), then \(\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1-x_{0}^{2}}}{x_{1} x_{2} x_{3} \ldots x_{n+1}}\right)\) is equal to (a) \(-1\) (b) 1 (c) \(\cos ^{-1}\left(x_{0}+1\right)\) (d) \(\cos ^{-1}\left(x_{0}\right)\)

For \(n \in N\), let \(f_{n}(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)(1+\sec 4 x) \ldots\left(1+\sec 2^{n} x\right)\) Then, \(\lim _{x \rightarrow 0} \frac{f_{n}(x)}{2 x}\) is equal to (a) 0 (b) \(2^{n}\) (c) \(2^{n-1}\) (d) \(2^{n+1}\)

If \(f(x)=\left\\{\begin{array}{cl}3+|x-k| & , \text { for } x \leq k \\\ a^{2}-2+\frac{\sin (x-k)}{x-k} & , \text { for } x>k\end{array}\right.\) has minimum at \(x=k\), then (a) \(a \in R\) (b) \(|a|<2\) (c) \(|a|>2\) (d) \(1<|a|<2\)

The vlaue of \(\lim _{x \rightarrow 0} \frac{\sqrt{\sec ^{2} \frac{x}{2}-1}}{x}\) is (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 1 (d) doesn't exist

Let \(f(x)\) be a real valued function defined for all \(x \geq 1\), satisfying \(f(1)=1\) and \(f^{\prime}(x)=\frac{1}{x^{2}+(f(x))^{2}}\); then \(\lim _{x \rightarrow \infty} f(x)\) (a) doesn't exist (b) exists and less than \(\frac{\pi}{4}\) (c) exists and less than \(1+\frac{\pi}{x}\) (d) exists and equal to 0

The quadratic equation whose roots are the minimum value of \(\sin ^{2} \theta-\sin \theta+\frac{1}{2}\) and \(\lim _{x \rightarrow \infty} \sqrt{(x+1)(x+2)}-x\) is (a) \(3 x^{2}-7 x+3=0\) (b) \(8 x^{2}-14 x+3=0\) (c) \(x^{2}-7 x+3=0\) (d) \(2 x^{2}-7 x+3=0\)

Let \([x]\) denotes the greatest integer function. Let \(g(x)=\frac{\sin \frac{\pi}{4}[x]}{[x]}\), then \(g\) is such that (a) it is continuous at \(x=\frac{3}{2}\) (b) it is continuous at \(x=2\) (c) it is not continuous at any point (d) it has its right limit at \(x=1\) as \(\frac{1}{2}\)

If \(x_{1}=\sqrt{3}\) and \(x_{n+1}=\frac{x_{n}}{1+\sqrt{1+x_{n}^{2}}}, \forall n \in N\), then \(\lim _{x \rightarrow \infty} 2^{n} x_{n}\) equal to (a) \(\frac{3}{2 \pi}\) (b) \(\frac{2}{3 \pi}\) (c) \(\frac{2 \pi}{3}\) (d) \(\frac{3 \pi}{2}\)

\(\lim _{x \rightarrow a^{-}} \frac{\sqrt{x-b}-\sqrt{a-b}}{\left(x^{2}-a^{2}\right)},(a>b)\) (a) \(\frac{1}{4 a}\) (b) \(\frac{1}{a \sqrt{a-b}}\) (c) \(\frac{1}{2 a \sqrt{a-b}}\) (d) \(\frac{1}{4 a \sqrt{a-b}}\)

\(\lim _{n \rightarrow \infty}\left(\sin ^{n} 1+\cos ^{n} 1\right)^{n}\) is equal to (a) \(\cot 1\) (b) \(\tan 1\) (c) \(\cos 1\) (d) \(\sin 1\)

If \(\lim _{x \rightarrow \infty} 4 x\left(\frac{\pi}{4}-\tan ^{-1} \frac{x+1}{x+2}\right)=y^{2}+4 y+5\), then \(y\) can be equal to (a) 1 (b) \(-1\) (c) \(-4\) (d) \(-3\)

\(\lim _{x \rightarrow 0} \frac{1-\cos \left(x^{2}\right)}{x^{3}\left(4^{x}-1\right)}\) is equal to (a) \(\frac{1}{2} \ln 2\) (b) \(\ln 2\) (c) \(\ln 4\) (d) \(1-\frac{1}{2} \ln \left(\frac{e^{2}}{4}\right)\)

If \(f(x)=e^{[\cot x]}\) where \([y]\) represents the greatest integer less than or equal to \(y\), then (a) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=1\) (b) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\frac{1}{e}\) (c) \(\lim _{x \rightarrow \frac{x^{-}}{2}} f(x)=\frac{1}{e}\) (d) \(\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=1\)

\(\lim _{x \rightarrow 0}\left[m \frac{\sin x}{x}\right]\) is equal to (where \(m \in I\) and [.] denotes greatest integer function.) (a) \(m\), if \(m \leq 0\) (b) \(m-1\), if \(m>0\) (c) \(m-1\), if \(m<0\) (d) \(m\), if \(m>0\)

If \(\lim _{x \rightarrow 0}\left(1+a x+b x^{2}\right)^{2 / x}=e^{3}\), then (a) \(a=3, b=0\) (b) \(a=\frac{3}{2}, b=1\) (c) \(a=\frac{3}{2}, b=4\) (d) \(a=2, b=3\)

Statement I \(\lim _{x \rightarrow a} f(x)\) exists \(=k\), but \(\lim _{x \rightarrow k} g(x)\) does not exist. If \(\lim _{x \rightarrow a} g(f(x))\) exists, then \(x=a\) is a point of extremum for \(y=f(x)\). If \(f(x)\) is non- linear. Statement II \(\lim _{x \rightarrow k} g(x)\) does not exist, but \(\lim _{x \rightarrow a} g(f(x))\) exists, \(f(x)\) will approach \(k\) when \(x \rightarrow a\) through only one side.

Statement I \(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^{2} \frac{x}{2}\right)}{x^{2}}=\pi\) Statement II \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

Let \(a_{n}=\underbrace{2.99 \ldots 9,}_{n \text { times }} n \in N\) Statement I \(\left[\lim _{n \rightarrow \infty} a_{n}\right]=\lim _{n \rightarrow \infty}\left[a_{n}\right],[\cdot]\) denotes the greatest integer function. Statement II \(\lim _{n \rightarrow \infty} a_{n}=3\)

If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\). \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\) is equal to (a) \(\frac{1}{e}\) (b) \(-\frac{1}{e}\) (c) \(e\) (d) \(-e\)

If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\) \(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{\frac{1}{x}}\) is equal to (a) \(e^{1 / 2}\) (b) \(e^{-1 / 2}\) (c) \(e^{1}\) (d) \(\frac{1}{e}\)

\(\operatorname{Let} f(x)=\lim _{n \rightarrow-}\left(\cos \sqrt{\frac{x}{n}}\right)^{n}, g(x)=\lim _{n \rightarrow-}(1+x+x \sqrt[n]{e})^{n} .\) Now, consider the function \(y=h(x)\), where \(h(x)=\tan ^{-1}\left(g^{-1} f^{-1}(x)\right)\) \(\lim _{x \rightarrow 0} \frac{\ln (f(x))}{\ln (g(x))}\) is equal to (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 0 (d) 1

Match the statements of Column I with values of Column II. Column I (A) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan ^{-1}(\tan x)\) (B) \(\lim _{n \rightarrow \infty}\left[\sum_{r=1}^{n} \frac{1}{2^{r}}\right]([.]\) denotes the greatest integer function) (C) \(\lim _{x \rightarrow-} \sec ^{-1}\left(\frac{x}{x+1}\right)\) (D) \(\lim _{x \rightarrow \frac{\pi}{9}} \frac{\cos x}{(1-\sin x)^{2 / 3}}\) Column II (p) 0 (q) Doesn't exist (r) \(-\frac{\pi}{2}\) (s) \(\frac{\pi}{2}\)

Match the statements of Column I with values of Column II. Column I (A) \(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{n+r}{n-r}}\) (B) \(\lim _{x \rightarrow-}\left[\frac{1}{\sqrt{n^{2}-1}}+\frac{1}{n^{2}-2^{2}}+. .+\frac{1}{\sqrt{n^{2}-(n-1)^{2}}}\right]\) (C) \(\lim _{x \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{1 / n}\) Column If (p) \(\frac{\pi}{2}\) (q) \(\frac{\pi}{2}+1\) (r) \(\pi\) (s) \(\frac{1}{e}\)