Chapter 12

If \(\alpha\) is the positive root of the equation, \(p(x)=x^{2}-x-2=0\), then \(\lim _{x \rightarrow \alpha^{+}} \frac{\sqrt{1-\cos (p(x))}}{x+\alpha-4}\) is equal to: [Sep. \(\left.\mathbf{0 5}, \mathbf{2 0 2 0}(\mathbf{I})\right]\) (a) \(\frac{3}{2}\) (b) \(\frac{3}{\sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)

\(\lim _{x \rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^{2}+x^{4}}-1\right) / x}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1} \quad\) [Sep. 05, 2020 (II)] (a) is equal to \(\sqrt{e}\) (b) is equal to 1 (c) is equal to 0 (d) does not exist

Let \([t]\) denote the greatest integer \(\leq t\). If for some \(\lambda \in \mathbf{R}-\\{0,1\\}, \lim _{x \rightarrow 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=L\), then \(L\) is equal to : [Sep. 03, 2020 (I)] (a) 1 (b) 2 (c) \(\frac{1}{2}\) (d) 0

If \(\lim _{x \rightarrow 0}\left\\{\frac{1}{x^{8}}\left(1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}\right)\right\\}=2^{-k}\), then the value of \(k\) is

\(\lim _{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}\) is equal to [NA Jan. 7, 2020 (I)]

Let \(f(x)=5-|x-2|\) and \(g(x)=|x+1|, x \in \mathrm{R}\). If \(f(x)\) attains maximum value at \(\alpha\) and \(g(x)\) attains minimum value at \(\beta\), then \(\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^{2}-5 x+6\right)}{x^{2}-6 x+8}\) is equal to : [April 12, 2019 (II)] (a) \(1 / 2\) (b) \(-3 / 2\) (c) \(-1 / 2\) (d) \(3 / 2\)

\(\lim _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^{2}+2 \sin x+1}-\sqrt{\sin ^{2} x-x+1}}\) is [April 12, 2019 (II)] (a) 6 (b) 2 (c) 3 (d) 1

If \(\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}=\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}\), then \(\mathrm{k}\) is: \([\) April \(\mathbf{1 0}, \mathbf{2 0 1 9}(\mathbf{I})]\) (a) \(\frac{8}{3}\) (b) \(\frac{3}{8}\) (c) \(\frac{3}{2}\) (d) \(\frac{4}{3}\)

If \(\lim _{x \rightarrow 1} \frac{x^{2}-a x+b}{x-1}=5\), then \(a+b\) is equal to : [April 10, 2019(II)] (a) \(-4\) (b) 5 (c) \(-7\) (d) 1

\(\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}\) equals: \(\quad\) [April 8, 2019 (I)] (a) \(4 \sqrt{2}\) (b) \(\sqrt{2}\) (c) \(2 \sqrt{2}\) (d) 4

\(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^{3} x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}\) is: [Jan. 12, 2019 (I)] (a) 4 (b) \(4 \sqrt{2}\) (c) \(8 \sqrt{2}\) (d) 8

\(\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}\) is equal to: \(\quad\) Jan. 12, 2019 (II)] (a) \(\frac{1}{\sqrt{2 \pi}}\) (b) \(\sqrt{\frac{2}{\pi}}\) (c) \(\sqrt{\frac{\pi}{2}}\) (d) \(\sqrt{\pi}\)

Let \([\mathrm{x}]\) denote the greatest integer less than or equal to \(\mathrm{x}\). Then : \(\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}}\) [Jan. 11, 2019(I)] (a) does not exist (b) equals \(\pi\) (c) equals \(\pi+1\) (d) equals 0

\(\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)}\) is equal to: \(\quad\) [Jan. 11, 2019 (II)] (a) 0 (b) 2 (c) 4 (d) 1

For each \(\mathrm{t} \in \mathbf{R}\), let \([\mathrm{t}]\) be the greatest integer less than or equal to t. Then, \(\lim _{x \rightarrow 1+} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}\) (a) equals 1 (b) equals 0 (c) equals \(-1\) (d) does not exist

\(\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}} \quad\) [Jan. 9, 2019 (I)] (a) exists and equals \(\frac{1}{4 \sqrt{2}}\) (b) exists and equals \(\frac{1}{2 \sqrt{2}(\sqrt{2}+1)}\) (c) exists and equals \(\frac{1}{2 \sqrt{2}}\) (d) does not exist

For each \(x \in \mathbf{R}\), let \([x]\) be greatest integer less than or equal to \(x\). Then \(\quad\) [Jan. 09, 2019 (II)] \(\lim _{x \rightarrow 0} \frac{x([x]+|x|) \sin [x]}{x}\) is equal to: (a) \(-\sin 1\) (b) 1 (c) \(\sin 1\) (d) 0

\(\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}\) equals. [Online April 15, 2018] (a) 1 (b) \(-\frac{1}{2}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{2}\)

\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^{3}}\) equals: [2017] (a) \(\frac{1}{4}\) (b) \(\frac{1}{24}\) (c) \(\frac{1}{16}\) (d) \(\frac{1}{8}\)

\(\lim _{x \rightarrow 3} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}}\) is equal to: \([\) Online April 8, 2017] (a) \(\sqrt{3}\) (b) \(\frac{1}{\sqrt{2}}\) (c) \(\frac{\sqrt{3}}{2}\) (d) \(\frac{1}{2 \sqrt{2}}\)

\(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to: \(\quad\) [2015] (a) 2 (b) \(\frac{1}{2}\) (c) 4 (d) 3

\(\lim _{x \rightarrow 0} \frac{e^{x^{2}}-\cos x}{\sin ^{2} x}\) is equal to: \(\quad\) [OnlineApril 10,2015] (a) 2 (b) 3 (c) \(\frac{3}{2}\) (d) \(\frac{5}{4}\)

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}\) is equal to: (a) \(-\pi\) (b) \(\pi\) (c) \(\frac{\pi}{2}\) (d) 1

\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+2}\right)^{x}\) (a) \(e^{4}\) (b) \(e^{2}\) (c) \(e^{3}\) (d) 1

If \(\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\\{x^{2}+(k-2) x-2 k\right\\}}{x^{2}-4 x+4}=5\), then \(\mathrm{k}\) is equal to: \(\quad\) [Online April 11, 2014] (a) 0 (b) 1 (c) 2 (d) 3

\(\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}\) is (a) 1 (b) \(-1\) (c) zero (d) does not exist

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

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}\) equals \(\quad\) [Online May 26, 2012] (a) \(-\pi\) (b) 1 (c) \(-1\) (d) \(\pi\)

\(\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)\) [Online May 7, 2012] (a) equals 1 (b) equals 0 (c) does not exist (d) equals \(-1\)

Let \(f: R \rightarrow[0, \infty)\) be such that \(\lim _{x \rightarrow 5} f(x)\) exists and \(\lim _{x \rightarrow 5} \frac{(f(x))^{2}-9}{\sqrt{|x-5|}}=0\) Then \(\lim _{x \rightarrow 5} f(x)\) equals: (a) 0 (b) 1 (c) 2 (d) 3

\(\lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \\{2(x-2)\\}}}{x-2}\right)\) (a) equals \(\sqrt{2}\) (b) equals \(-\sqrt{2}\) (c) equals \(\frac{1}{\sqrt{2}}\) (d) does not exist

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a positive increasing function with \(\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1\) then \(\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=\) [2010] (a) \(\frac{2}{3}\) (b) \(\frac{3}{2}\) (c) 3 (d) 1

Let \(\alpha\) and \(\beta\) be the distinct roots of \(a x^{2}+b x+c=0\), then \(\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^{2}+b x+c\right)}{(x-\alpha)^{2}}\) is equal to \([2005]\) (a) \(\frac{a^{2}}{2}(\alpha-\beta)^{2}\) (b) 0 (c) \(\frac{-a^{2}}{2}(\alpha-\beta)^{2}\) (d) \(\frac{1}{2}(\alpha-\beta)^{2}\)

\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[1-\tan \left(\frac{x}{2}\right)\right][1-\sin x]}{\left[1+\tan \left(\frac{x}{2}\right)\right][\pi-2 x]^{3}}\) is \(\quad\) [2003] (a) \(\infty\) (b) \(\frac{1}{8}\) (c) 0 (d) \(\frac{1}{32}\)

\(\lim _{x \rightarrow 0} \frac{\log x^{n}-[x]}{[x]}, n \in N,([x]\) denotes greatest integer less than or equal to \(x\) ) [2002] (a) has value \(-1\) (b) has value 0 (c) has value 1 (d) does not exist

\(\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0)\) is equal to : [Sep. 03, 2020 (II)] (a) \(\left(\frac{2}{3}\right)^{\frac{4}{3}}\) (b) \(\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}\) (c) \(\left(\frac{2}{9}\right)^{\frac{4}{3}}\) (d) \(\left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}}\)

If \(\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots+x^{n}-n}{x-1}=820,(n \in \mathbf{N})\) then the value of \(n\) is equal to

\(\lim _{x \rightarrow 0}\left(\tan \left(\frac{\pi}{4}+x\right)\right)^{1 / x}\) is equal to: \(\quad\) [Sep. \(\mathbf{0 2}, \mathbf{2 0 2 0}\) (II)] (a) \(e\) (b) 2 (c) 1 (d) \(e^{2}\)

\(\lim _{x \rightarrow 0}\left(\frac{3 x^{2}+2}{7 x^{2}+2}\right)^{1 / x^{2}}\) is equal to: \(\quad\) [Jan. 8, 2020 (I)] (a) \(\frac{1}{e}\) (b) \(\frac{1}{e^{2}}\) (c) \(\mathrm{e}^{2}\) (d) \(\mathrm{e}\)

\(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{t \sin (10 t) d t}{x}\) is equal to: [Jan. \(8,2020(\mathrm{II})]\) (a) 0 (b) \(\frac{1}{10}\) (c) \(-\frac{1}{5}\) (d) \(-\frac{1}{10}\)

If \(\alpha\) and \(\beta\) are the roots of the equation \(375 x^{2}-25 x-2=0\) then \(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \alpha^{r}+\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \beta^{r}\) is equal to : [April 12, 2019 (I)] (a) \(\frac{21}{346}\) (b) \(\frac{29}{358}\) (c) \(\frac{1}{12}\) (d) \(\frac{7}{116}\)

Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function satisfying \(f^{\prime}(3)+f^{\prime}(2)=0\). Then \(\lim _{x \rightarrow 0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}}\) is equal to : \(\quad\) [April 08,2019 (II)] (a) 1 (b) \(e^{-1}\) (c) \(e\) (d) \(e^{2}\)

For each \(t \in R\), let \([t]\) be the greatest integer less than or equal to \(\mathrm{t}\). Then \([2018]\) \(\lim _{x \rightarrow 0^{+}} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{15}{x}\right]\right)\) (a) is equal to 15 . (b) is equal to 120 . (c) does not exist (in \(\mathrm{R}\) ). (d) is equal to 0 .

\(\lim _{x \rightarrow 0} \frac{(27+x)^{3}-3}{9-(27+x)^{\frac{2}{3}}}\) equals. [Online April 16, 2018] (a) \(-\frac{1}{3}\) (b) \(\frac{1}{6}\) (c) \(-\frac{1}{6}\) (d) \(\frac{1}{3}\)

Let \(\mathrm{p}=\lim _{\mathrm{x} \rightarrow 0^{+}}\left(1+\tan ^{2} \sqrt{\mathrm{x}}\right)^{\frac{1}{2 \mathrm{x}}}\) then \(\log \mathrm{p}\) is equal to : \([2016]\) (a) \(\frac{1}{2}\) (b) \(\frac{1}{4}\) (c) 2 (d) 1

\(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)^{2}}{2 x \tan x-x \tan 2 x}\) is : \(\quad\) [Online April 10,2016] (a) 2 (b) \(-\frac{1}{2}\) (c) \(-2\) (d) \(\frac{1}{2}\)

If \(\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}-\frac{4}{x^{2}}\right)^{2 x}=e^{3}\), then 'a' is equal to : [Online April 9, 2016] (a) 2 (b) \(\frac{3}{2}\) (c) \(\frac{1}{2}\) (d) \(\frac{2}{3}\)

If \(\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^{2}}\right)^{2 x}=e^{2}\), then the values of \(a\) and \(b\), are [2004] (a) \(a=1\) and \(b=2\) (b) \(a=1, b \in \mathrm{R}\) (c) \(a \in \mathrm{R}, b=2\) (d) \(a \in \mathrm{R}, b \in \mathrm{R}\)

Let \(f(x)\) be a polynomial of degree 4 having extreme values at \(x=1\) and \(x=2\). If \(\lim _{x \rightarrow 0}\left(\frac{f(x)}{x^{2}}+1\right)=3\) then \(f(-1)\) is equal to [Online April 15, 2018] (a) \(\frac{1}{2}\) (b) \(\frac{3}{2}\) (c) \(\frac{5}{2}\) (d) \(\frac{9}{2}\)

Let \(\mathrm{f}(\mathrm{x})\) be a polynomial of degree four having extreme values at \(\mathrm{x}=1\) and \(\mathrm{x}=\) 2. If \(\lim _{\mathrm{x} \rightarrow 0}\left[1+\frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{2}}\right]=3\), then \(\mathrm{f}(2)\) is equal to : (a) 0 (b) 4 (c) \(-8\) (d) \(-4\)