Limits, Continuity and Differentiability

$ \begin{aligned} & \text { } \lim _{x \rightarrow \infty}[x-\log (\cosh x)] \\ & \because \quad \cosh x=\frac{e^x+e^{-x}}{2} \end{aligned} $

$\because x \rightarrow \infty, e^{-x}$ become negligible

$ \begin{aligned} \therefore \cosh x & \simeq \frac{e^x}{2} \\ \Rightarrow \log (\cosh x) & =\log \left(\frac{e^x}{2}\right)=x-\log 2 \\ & =\lim _{x \rightarrow \infty}[x-x+\log 2]=\log 2 \end{aligned} $

$ \begin{aligned} & \left.\text { } \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)\right. \\ & =\lim _{x \rightarrow \infty}\left(x\left(\sqrt[3]{1+\frac{4}{x}}-\sqrt{1-\frac{3}{x}}\right)\right) \end{aligned} $

$ \begin{aligned} &\text { Apply binomial expansion for radicals }\\ &\begin{aligned} & =\lim _{x \rightarrow \infty}\left(x \left(\left(1+\frac{1}{3} \cdot \frac{4}{x}+0\left(\frac{1}{x^2}\right)\right)\right.\right. \left.-\left(1-\frac{1}{2} \cdot \frac{3}{x}+0\left(\frac{1}{x^2}\right)\right)\right) \\ & =\lim _{x \rightarrow \infty} x\left(1+\frac{4}{3 x}-1+\frac{3}{2 x}\right) \\ & =\lim _{x \rightarrow \infty} x \cdot \frac{1}{x}\left(\frac{4}{3}+\frac{3}{2}\right) \\ & =\frac{4}{3}+\frac{3}{2}=\frac{8+9}{6}=17 / 6 \end{aligned} \end{aligned} $

$\because f(x)$ is continuous at $x=1$

$ \therefore \mathop {\lim }\limits_{x \to {1^ - }} f(x)=f(1)=\mathop {\lim }\limits_{x \to {1^ +}} f(x) $

So, $\mathop {\lim }\limits_{x \to {1^ - }} f(x)=\mathop {\lim }\limits_{x \to {1^ - }} \left[e^{\frac{\sin a(x-[x])}{x-[x]}}\right]$

$ =\mathop {\lim }\limits_{x \to {1^ - }} \left(e^{\frac{\sin a(x-0)}{x-0}}\right)=e^{\sin a} $

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x)=\mathop {\lim }\limits_{x \to {1^+ }} \frac{\left|x^2+x-2\right|}{x-1}$

$ =\mathop {\lim }\limits_{x \to {1^+ }} \frac{(x-1)(x+2)}{(x-1)}=\mathop {\lim }\limits_{x \to {1^+ }}(x+2)=3 $

and $\quad f(1)=b+1=3$

$\Rightarrow b=2$ and $e^{\sin a}=3 \Rightarrow \sin a=\ln 3$

Hence, $b \sin a=2 \ln 3=\ln 9$

We have,

$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}\left(\frac{0}{0} \text { form }\right) $

Using L Hospital rule,

$ \begin{aligned} & \lim _{x \rightarrow \frac{\pi}{4}} \frac{-3(\cos x+\sin x)^2(\cos x-\sin x)}{-2 \cos 2 x} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-3(\cos x+\sin x)\left(\cos ^2 x-\sin ^2 x\right)}{-2(\cos 2 x)} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{3(\cos x+\sin x)(\cos 2 x)}{2 \cos 2 x} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{3}{2}(\cos x+\sin x) \\ & =\frac{3}{2} \times \sqrt{2}=\frac{3}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) \\ & \lim _{x \rightarrow 2^{+}}\left(\frac{\left[2^{+}\right]^3}{3}-\left[\frac{2^{+}}{3}\right]^3\right)=\frac{8}{3}-0=\frac{8}{3} \end{aligned} \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{cl} \frac{1-\cos 4 x}{x^2}, & x<0 \\ a & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{array}\right. $

$\because f(x)$ is continuous at $x=0$

$ \begin{aligned} \therefore f(0) & =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ a & =\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\ \Rightarrow a & =\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4}{16+\sqrt{x}-16} \\ \Rightarrow \quad a & =\lim _{x \rightarrow 0^{-}} 8\left(\frac{\sin 2 x}{2 x}\right)^2 \\ & =\lim _{x \rightarrow 0^{+}} \sqrt{16+\sqrt{x}}+4 \\ \quad a & =8=8 \\ \Rightarrow \quad a & =8 \end{aligned} $

$ \text { } \begin{aligned} f(x) & =\frac{x}{1+|x|} \\ & =\left[\begin{array}{ll} \frac{x}{1-x}, & x<0 \\ \frac{x}{1+x}, & x \geq 0 \end{array}\right. \end{aligned} $

∵ Given function is differentiable for all real $x$ but we have to check at $x=0$

$ \begin{aligned} & \begin{aligned} \text { LHD } & =f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{\frac{-h}{1+h}-0}{-h}=1 \end{aligned} \\ & \text { RHD } =f^{\prime}\left(0^{+}\right) =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & \Rightarrow \lim _{h \rightarrow 0} \frac{\frac{h}{1+h}-0}{h}=1 \end{aligned} $

$ \because L H D=R H D $

∴ Differentiable at $x=0$

$\therefore f(x)$ is differentiate for $x \in(-\infty, \infty)$

$ \mathop {\lim }\limits_{x \to 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3-\sqrt{\sin ^2 x-2 \tan x-x+3}}} $

$ \begin{aligned} &\text { After rationalisation }\\ &\left(x+2 \sin x+3 \tan x-\tan ^3 x\right) \end{aligned} $

$ \mathop {\lim }\limits_{x \to 0} \frac{\left(\sqrt{x^2+2 \sin x+\tan x+3}\right.+\sqrt{\sin ^2 x-2 \tan x-x+3}}{x^2+2 \sin x+\tan x+3-\sin ^2 x+2 \tan x+x-3} $

$ x\left(1+\frac{2 \sin x}{x}+\frac{3 \tan x}{x}-\frac{3 \tan ^3 x}{x}\right) $

$ \mathop {\lim }\limits_{x \to 0} \frac{\left.\left(\sqrt{x^2+2 \sin x+\tan x+3}\right.+\sqrt{\sin ^2 x-2 \tan x-x+3}\right)}{x\left(1+x+\frac{2 \sin x}{x}+\frac{\tan x}{x}\right.\left.-\frac{\sin ^2 x}{x}+\frac{2 \tan x}{x}\right)} $

$ \Rightarrow \frac{(1+2+3)(\sqrt{3}+\sqrt{3})}{(1+2+1+2)}=\frac{6 \times 2 \sqrt{3}}{6}=2 \sqrt{3} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{x \rightarrow \infty} \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}} \\ & =\lim _{x \rightarrow \infty} \frac{x^{25} \cdot x^{35}\left(\frac{3}{x}-1\right)^{25}\left(\frac{6}{x}+1\right)^{35}}{x^{38} \cdot x^{22}\left(\frac{12}{x}+1\right)^{38}\left(\frac{9}{x}-1\right)^{22}} \\ & =\lim _{x \rightarrow \infty} \frac{\left(\frac{3}{x}-1\right)^{25}\left(\frac{6}{x}+1\right)^{35}}{\left(\frac{12}{x}+1\right)^{38}\left(\frac{9}{x}-1\right)^{22}} \end{aligned} \end{aligned} $

$ =\frac{(-1)(1)}{(1)(1)}=-1\left\{\begin{array}{c} x \rightarrow \infty \\ \frac{3}{x} \rightarrow 0 \\\because \frac{6}{x} \rightarrow 0 \\ \frac{12}{x} \rightarrow 0 \\ \frac{9}{x} \rightarrow 0 \end{array}\right\} $

We have,

$ f(x)=\left\{\begin{array}{cc} \log (1+[x]) & x \geq 0 \\ \sin ^{-1}[x] & -1 \leq x<0 \\ k([x]+|x|] & x<-1 \end{array}\right. $

$\because f(x)$ is continuous at $x=-1$

$ \begin{aligned} & \therefore f(-1)=f\left(-1^{-}\right)=f\left(-1^{+}\right) \\ & \sin ^{-1}(-1)=k\left(\left[-1^{-}\right]+\left|-1^{-}\right|\right)=\sin ^{-1}\left[-1^{+}\right] \\ & \Rightarrow \quad \frac{-\pi}{2}=k(-2+1)=\frac{-\pi}{2} \\ & \Rightarrow \quad-k=-\pi / 2 \\ & \therefore \quad k=\frac{\pi}{2} \end{aligned} $

$ \text { } \begin{aligned} \lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}\right. & +\sin \frac{3 \pi}{2 n} \left.+\ldots+\sin \frac{\pi}{2}\right] \\ = & \lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left(\frac{\sin \frac{n}{2}\left(\frac{\pi}{2 n}\right)}{\sin \frac{1}{2}\left(\frac{\pi}{2 n}\right)} \cdot \sin \left(\frac{\frac{\pi}{2 n}+\frac{\pi}{2}}{2}\right)\right) \\ = & \lim _{n \rightarrow \infty} \frac{\pi}{2 n} \cdot \frac{\left(\sin \frac{\pi}{4}\right)}{\sin \frac{\pi}{4 n}} \cdot \sin \left(\frac{\pi}{4 n}+\frac{\pi}{4}\right) \end{aligned} $

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \frac{(1 / \sqrt{2})}{\left(\frac{\sin \frac{\pi}{4 n}}{\frac{\pi}{4 n}}\right)} \cdot \sin \left(\frac{\pi}{4 n}+\frac{\pi}{4}\right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =2 \cdot \frac{1}{\sqrt{2}} \sin \pi / 4=1 \end{aligned} $

Since, $x \rightarrow 0^{+}$, so, the greatest integer function $[x]$ equals 0 as $0 \leq x<1$.

$ \Rightarrow \quad \lim\limits_{x \rightarrow 0^{+}} \frac{\cos (0)-\cos (k x-0)}{x^2}=5 \Rightarrow $

$ \begin{aligned} & \lim\limits _{x \rightarrow 0^{+}} \frac{1-\cos (k x)}{x^2}=5 \\ & \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{2 \sin ^2\left(\frac{k x}{2}\right)}{x^2}=5 \\ & \Rightarrow \lim _{x \rightarrow 0^{+}} 2 \frac{\sin ^2\left(\frac{k x}{2}\right)}{\left(\frac{k x}{2}\right)^2} \cdot \frac{k^2}{4}=5 \\ & \Rightarrow \quad 2 \cdot 1 \cdot \frac{k^2}{4}=5 \quad\left[\because \lim _{\theta \rightarrow 0} \frac{\sin (\theta)}{\theta}=1\right] \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad k^2=\frac{20}{2} \Rightarrow k= \pm \sqrt{10}\\ &\text { So, } k=\sqrt{10} \end{aligned} $

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

$ \begin{array}{ll} =\frac{1}{2 \cos ^3 0\left(1-\tan ^2 0\right)} \cdot 1 & {\left[\because \lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right]} \\ =\frac{1}{2(1)(1)}=\frac{1}{2} & \end{array} $

Given, function

$ f(x)=\left\{\begin{array}{cc} \frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x} & , \quad x>0 \\ \frac{\cos 4 x-\cos b x}{\tan ^2 x} & , \quad x=0 \end{array}\right. $

is contiuous at $x=0$

Since, $f(x)$ is continuous at $x=0$

So, $\mathop {\lim }\limits_{x \to 0 - } f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$ and $f(0)=2$

Now, $\mathop {\lim }\limits_{x \to 0 + } \frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x}$

Applying L'hospital rule, twice, since it is an indeterminate form $\frac{0}{0}$.

So, $\mathop {\lim }\limits_{x \to 0 + } \frac{a e^{a x} \log (1+x)+\frac{\left(e^{a x}-1\right)}{1+x}}{2 \cos x \sin x}$

This limit is again $\frac{0}{0}$ form, so apply L'Hospital rule again.

$\mathop {\lim }\limits_{x \to 0 + } \frac{a^2 e^{a x} \log (1+x)+\frac{a e^{a x}(1+x)-\left(e^{a x}-1\right)}{(1+x)^2}+\frac{a e^{a x}}{1+x}}{2 \cos ^2 x-2 \sin ^2 x}$

$ =\frac{0+\frac{a(1)-(1-1)}{1^2}+\frac{a}{1}}{2(1)-0}=\frac{a+a}{2}=a $

Since, $\mathop {\lim }\limits_{x \to 0 + } f(x)=f(0) \Rightarrow a=2$

Now, $\mathop {\lim }\limits_{x \to 0 - } \frac{\cos 4 x-\cos b x}{\tan ^2 x}$

$ =\mathop {\lim }\limits_{x \to 0 - } \frac{-2 \sin \left(\frac{4 x+b x}{2}\right) \sin \left(\frac{4 x-b x}{2}\right)}{\tan ^2 x} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{-2 \sin \left(\frac{4 x+b x}{2}\right) \sin \left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{\frac{\sin ^2 x}{x^2} \cdot x^2} \\ & \lim _{x \rightarrow 0^{-}} \frac{\frac{-2 \sin \left(\frac{4 x+b x}{2}\right)}{\frac{4 x+b x}{2}} \cdot\left(\frac{4 x+b x}{2}\right) \cdot \frac{\sin \left(\frac{4 x-b x}{2}\right)}{\left(\frac{4 x-b x}{2}\right)}\left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{\frac{\sin ^2 x}{x^2} \cdot x^2} \\ & =\lim _{x \rightarrow 0^{-}} \frac{-2 \cdot 1 \cdot\left(\frac{4 x+b x}{2}\right) \cdot 1 \cdot\left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{1 \cdot x^2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ & =\lim _{x \rightarrow 0^{-}} \frac{-2\left(16 x^2-b^2 x^2\right)}{4 x^2} \cdot \cos ^2 x \\ & =\lim _{x \rightarrow 0^{-}} \frac{\left(b^2-16\right)}{2} \cdot \cos ^2 x=\frac{b^2-16}{2} \cdot 1 \quad\left[\because \lim _{x \rightarrow 0^{-}} \cos ^2 x=1\right] \end{aligned} $

$ =\frac{b^2-16}{2} $

Also, $\mathop {\lim }\limits_{x \to 0 - } f(x)=f(0)$

$ \Rightarrow \quad \frac{b^2-16}{2}=2 \Rightarrow b^2=4+16=20 $

Now, $\sqrt{b^2-a^2}=\sqrt{20-2^2}=\sqrt{20-4}=\sqrt{16}=4$

$ \text { } \begin{aligned} & \lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2} \\ & =\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{\left(2 \sin ^2\left(\frac{3 x}{2}\right)\right)^2} \end{aligned} $

For small-angles $\{\because x \rightarrow 0\}$

$ \begin{aligned} & \sin k x \rightarrow k x \\ & =\lim _{x \rightarrow 0} \frac{x^2(3 x)^2+(6 x)^4}{\left(2\left(\frac{3 x}{2}\right)^2\right)^2} \\ & =\lim _{x \rightarrow 0} \frac{x^4[9+1296]}{x^4\left[\frac{9}{2}\right]^2}=\lim _{x \rightarrow 0} \frac{1305}{81} \times 4 \\ & =\frac{435}{27} \times 4=\frac{580}{9} \end{aligned} $

$ f(x)=\left\{\begin{array}{cc} (1+\sin x)^{\cos x}, & -\frac{\pi}{2} < x < 0 \\ a, & x=0 \\ \frac{e^{\frac{2}{x}}+e^{\frac{3}{x}}}{a e^{\frac{2}{x}}+b e^{\frac{3}{x}}}, & 0 < x < \frac{\pi}{2} \end{array}\right. $

So, $\mathop {\lim }\limits_{x \to {0^ - }} f(x)=\lim _{x \rightarrow 0^{-}}(1+\sin x)^{\cos x}$

Let $y=(1+\sin x)^{\cos x}$

$ \begin{aligned} \Rightarrow \ln y & =\cos x \ln (1+\sin x) \\ & =\frac{\ln (1+\sin x)}{\sin x}=1 \end{aligned} $

$\Rightarrow \quad y=e$

$ \therefore \mathop {\lim }\limits_{x \to {0^ - }}(1+\sin x)^{\cos x}=e=a=f(0) $

and $\mathop {\lim }\limits_{x \to {0^ +}} f(x)=\mathop {\lim }\limits_{x \to {0^+ }} \frac{e^{\frac{2}{x}}+e^{\frac{3}{x}}}{a e^{\frac{2}{x}}+b e^{\frac{3}{2}}}$

$ =\mathop {\lim }\limits_{x \to {0^ + }} \frac{e^{\frac{3}{x}}\left(e^{-\frac{1}{x}}+1\right)}{e^{\frac{3}{x}}\left(e \cdot e^{\frac{-1}{x}}+b\right)}=\frac{1}{b} $

So, $e=\frac{1}{b} \Rightarrow b=\frac{1}{e}$

$\mathop {\lim }\limits_{x \to 0} \frac{(\operatorname{cosec} x-\cot x)\left(e^x-e^{-x}\right)}{\sqrt{3}-\sqrt{2+\cos x}}$

Convert, $\operatorname{cosec} x$ and $\cot x$ into $\sin x$ and $e^x-e^{-x}=2 \sin h x$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\left(\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}} \cdot 2 \sin h x\right)}{\sqrt{3}-\sqrt{2+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x}{\sqrt{3}-\sqrt{2+\cos x}} \times \frac{\sqrt{3}+\sqrt{2+\cos x}}{\sqrt{3}+\sqrt{2+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{2 \sin \frac{x}{2}}{\cos \frac{x}{2}} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{2 \sin ^2 \frac{x}{2}} \\ & =\lim _{x \rightarrow 0} \frac{\sin h x(\sqrt{3}+\sqrt{2+\cos x}}{\cos \frac{x}{2} \cdot \sin \frac{x}{2}} \\ & \because x \rightarrow 0 \Rightarrow \cos \frac{x}{2} \rightarrow 1 \text { and } \sin \frac{x}{2} \rightarrow \frac{x}{2} \\ & \sin h x \rightarrow x \\ & \quad=\frac{x(\sqrt{3}+\sqrt{2+1})}{\frac{x}{2}}=2(\sqrt{3}+\sqrt{3}) \\ & \quad=2(2 \sqrt{3})=4 \sqrt{3} \end{aligned} $

$l=\mathop {\lim }\limits_{\theta \to 0} 3\left(\frac{\sin \theta}{\theta}\right)-\mathop {\lim }\limits_{\theta \to 0} 4 \cdot\left(\frac{\sin \theta}{\theta}\right)^3 \cdot \theta^2=3 \times 1-4 \times 0=3$

$ m=\mathop {\lim }\limits_{\theta \to 0}\left(\frac{2 \tan \theta}{\theta} \times \frac{1}{1-\tan ^2 \theta}\right)=2 \times 1 \times\left(\frac{1}{1-0}\right)=2 $

$\therefore$ The quadratic equation whose roots are 3 and 2 is

$ x^2-5 x+6=0 . $

$ \begin{aligned} & \text { Let } L=\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}} \\ & \left.=\lim _{x \rightarrow \infty} \frac{x\left(3+\frac{4 \cos ^2 x}{x}\right)}{x \sqrt{1-5\left(\frac{\sin x}{x}\right)^2}}\left(\because \lim _{x \rightarrow \infty}\left(\frac{\sin x}{x}\right)^2=\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{1}{y}\right)}{\frac{1}{y}}\right)\right)^2\right) \\ & =\lim _{y \rightarrow 0}\left(y \sin \left(\frac{1}{y}\right)\right)^2=(0 \times a \text { finite quantity })=0=\frac{3+0}{\sqrt{1-0}}=3 \end{aligned} $

LHL

$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{\left(1+a x^2+b x^3\right)^{\frac{1}{3}}-\left(1-a x^2-b x^3\right)^{\frac{1}{3}}}{x^2}\left(\text { form } \frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0^{-}} \frac{\frac{1}{3}\left(1+a x+b x^3\right)^{\frac{1}{3}-1}\left(2 a x+3 b x^2\right)-\frac{1}{3}\left(1-a x^2-b x^3\right)^{\frac{1}{3}-1}(-2 a x-3 b x)}{2 x} \end{aligned} $

(Using L Hospital rule)

$ =\frac{\frac{1}{3} \times(1+0+0)^{\frac{-2}{3}}(2 a+0)-\frac{1}{3}(1-0-0)^{\frac{-2}{3}}(-2 a-0)}{2} $

(cancel $x$ from both $N^r \times \operatorname{Dr}$ )

$ =\frac{\frac{2 a}{3}+\frac{2 a}{3}}{2}=\frac{4 a}{6}=\frac{2 a}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} &\text { RHL }\\ &\left(3 x+\frac{(3 x)^3}{3}+\frac{2(3 x)^5}{15}+\ldots\right) \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x-\sin 3 x}{b x^3}=\lim _{x \rightarrow 0^{+}} \frac{-\left(3 x-\frac{(3 x)^3}{3!}+\ldots\right)}{b x^3} \\ & \quad=\lim _{x \rightarrow 0^{+}} \frac{\frac{27 x^3}{3}+\frac{27 x^3}{6}+\text { (higher terms are zero) }}{b x^3}=\frac{27}{2 b} \\ & f(0)=5 \\ & \because \quad f(x) \text { is continuous at } x=0 . \\ & \therefore \quad \frac{2 a}{3}=5=\frac{27}{2 b} \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad a=\frac{15}{2}, b=\frac{27}{10}\\ &\therefore \text { Geometric mean of } a \text { and } b=\sqrt{a b}=\sqrt{\frac{15}{2} \times \frac{27}{10}}=\frac{9}{2} \end{aligned} $

Evaluate $[x]$ as $x \rightarrow 0^{-}$

As $x$ approaches 0 from the left $[x]=-1 \{x\}$ as $x \rightarrow 0^{-}$

$ \{x\}=x-[x]=x-(-1)=x+1 $

as $x \rightarrow 0^{-},\{x\} \rightarrow 1$

$ \begin{aligned} \lim _{x \rightarrow 0^{-}} & \frac{\sin ^{-1}(x+[x])}{2-\{x\}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{2-(x+1)} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{1-x} \end{aligned} $

as $x \rightarrow 0^{-}$, the limit becomes

$ \frac{\sin ^{-1}(-1)}{1-0}=-\frac{\pi}{2}=-\frac{\pi}{2} $

Therefore, $\theta=-\frac{\pi}{2}$

$ \begin{aligned} \sin \theta+\cos \theta & =\sin \left(-\frac{\pi}{2}\right)+\cos \left(-\frac{\pi}{2}\right) \\ & =-1 \\ \sin \theta+\cos \theta & =-1 \end{aligned} $

$ \text { } \begin{aligned} & \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x \\ = & \lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1) x}{6} \\ = & \lim _{n \rightarrow \infty} \frac{n \cdot n \cdot n\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) x}{6 n^3} \\ = & \frac{x}{6} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) \\ = & \frac{x}{6} \times 2=\frac{x}{3} \end{aligned} $

Since, $f(x)$ is continuous at $x=1$

$ \begin{array}{r} \Rightarrow \quad \mathop {\lim }\limits_{x \to {1^ - }} f(x)=f(1)=1 \\ =\mathop {\lim }\limits_{x \to {1^ + }} f(x) \end{array} $

Now,

$\mathop {\lim }\limits_{x \to {1^ - }} f(x)=\mathop {\lim }\limits_{x \to {1^ - }} b-\left\{\frac{\sin [x-1]-[x-1]}{[x-1]^3}\right\} $

Put $x-1=t$

$ \begin{aligned} & =\mathop {\lim }\limits_{x \to {1^ - }} b-\left\{\frac{\sin [t]-[t]}{[t]^3}\right\} \\ & =b-\left\{\frac{\sin (-1)-(-1)}{(-1)^3}\right\} \end{aligned} $

$ =b+(-\sin 1+1)=b-\sin 1+1 $

and

$ \mathop {\lim }\limits_{x \to {1^ +}} f(x)=\mathop {\lim }\limits_{x \to {1^+ }}\left(a-\frac{\sin [x-1]}{x-1}\right) $

Put $x-1=t$

$ =\mathop {\lim }\limits_{x \to {1^ + }}\left(a-\frac{\sin [t]}{t}\right)=a-1 $

By Eq. (i), $a=2$ and $b-\operatorname{sinl}+1=1$

$ \begin{array}{ll} \Rightarrow & a=2 \text { and } b=\sin 1 \\ \Rightarrow & a+b=2+\sin 1 \end{array} $

$\mathop {\lim }\limits_{y \to 0} \frac{\left(1+\sqrt{1+y^4}\right)^{\frac{1}{2}}-\sqrt{2}}{y^4}$

Using L'Hospital's rule,

$ \begin{aligned} & \lim _{y \rightarrow 0} \frac{\frac{1}{2}\left[1+\sqrt{1+y^4}\right]^{\frac{-1}{2}}\left(\frac{1}{2}\left(1+y^4\right)^{\frac{-1}{2}} 4 y^3\right)}{4 y^3} \\ & \Rightarrow \frac{1}{2} \frac{1}{(\sqrt{2})}\left(\frac{1}{2}\right)=\frac{1}{4 \sqrt{2}} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{\cos 2 x-\cos 4 x}{1-\cos 2 x} \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{2 \sin 3 x \sin x}{2 \sin ^2 x}=k \end{aligned} \end{aligned} $

$ \begin{aligned} &\lim _{x \rightarrow 0} \frac{3 \sin x-4 \sin ^3 x}{\sin x}=3=k\\ &\text { Now, } \lim _{x \rightarrow 3}\left(\frac{x^3-27}{x^4-81}\right) \Rightarrow \lim _{x \rightarrow 3} \frac{3 x^2}{4 x^3}\\ &=\frac{3(3)^2}{4(3)^3}=\frac{1}{4} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=\left\{\begin{array}{cc} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{array}\right. \\ & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ & 2=a \Rightarrow a=2 \\ & \lim _{x \rightarrow 2^{-}} f(x)=f(2)=\lim _{x \rightarrow 2^{+}} f(x) \\ & \Rightarrow \quad a-2=4-b^2 \Rightarrow b^2=4 \\ & a^2+b^2=8 \end{aligned} \end{aligned} $

$\mathop {\lim }\limits_{x \to - \infty } \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}$

$ \therefore \quad|x|=-x $

Now, dividing $N^t$ and $D^t$ by $x^3$, we get

$ \mathop {\lim }\limits_{x \to - \infty } \frac{5-\frac{\sin 5 x}{x}}{\frac{\cos 4 x}{x^2}-7+\frac{4}{x^2}+\frac{3}{x^3}} $

Clearly, $\mathop {\lim }\limits_{x \to - \infty } \frac{\sin 5 x}{x}=0$

and $\mathop {\lim }\limits_{x \to - \infty } \frac{\cos 4 x}{x^2}=0$

$ \begin{gathered} \therefore \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} \\ =\frac{5}{-7}=-\frac{5}{7} \end{gathered} $

$\mathop {\lim }\limits_{x \to {a^ + }} f(x)=p, \mathop {\lim }\limits_{x \to {a^ - }} f(x)=m$

and $\quad f(a)=k$

$ \begin{array}{lc} & \mathrm{LHL}=\mathrm{RHL}=f(a) \\ \therefore & p=m=k \end{array} $

$p-m=0$ and $p-k=0$, then $f(x)$ is right continuous at $x=a$.

$ \begin{aligned} &\text { We have, } f(0)=a\\ &\begin{aligned} & \text { LHL }=\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2} \times \frac{16}{16} \\ &=\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{(4 x)^2} \times 16=\frac{1}{2} \times 16=8 \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}} \\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x}[\sqrt{16+\sqrt{x}}+4]}{16+\sqrt{x}-16} \\ &=\lim _{x \rightarrow 0} \sqrt{16+\sqrt{x}}+4=8 \\ & \therefore \quad a=8 \end{aligned} \end{aligned} $

$\mathop {\lim }\limits_{x \to \infty } \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2 x-4}}$

By rationalizing $N^r$ and $D^r$, we get

$ \begin{aligned} \lim _{x \rightarrow 0} & \frac{2-1-\cos x}{\sqrt{2}+\sqrt{1+\cos x}} \times \frac{\sqrt{15+\cos 2 x}+4}{15+\cos 2 x-16} \\ & =\lim _{x \rightarrow 0} \frac{1-\cos x}{-(1-\cos 2 x)} \times \frac{4+4}{\sqrt{2}+\sqrt{2}} \\ & =-\frac{8}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{\frac{1-\cos x}{x^2} \times x^2}{1-\frac{\cos 2 x}{(2 x)^2} \times 4 x^2} \end{aligned} $

Using result $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}$

$ =-\frac{8}{2 \sqrt{2}} \times \frac{\frac{1}{2}}{\frac{1}{2}} \times \frac{1}{4}=-\frac{1}{\sqrt{2}} $

$\mathop {\lim }\limits_{x \to 3} \frac{x^2+(a+3) x+(a+1)}{x+3}$

It is $\frac{0}{0}$ form

$ \mathop {\lim }\limits_{x \to 3} \frac{2 x+(a+3)}{1}=-6+a+3=a-3 $

$\because f(x)$ is continuous $a-3=-\frac{5}{2}$

$ \begin{aligned} & a=3-\frac{5}{2}=\frac{1}{2} \\ & \lim _{x \rightarrow \frac{1}{2}}\left(x^2+x+1\right)=\frac{1}{4}+\frac{1}{2}+1=\frac{1}{4}+\frac{3}{2} \\ & =\frac{1+6}{4}=\frac{7}{4} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2} \\ & =\lim _{x \rightarrow 0} \frac{2 x \tan x\left[\frac{1}{1-\tan ^2 x}-1\right]}{\frac{1-\cos 3 x}{(3 x)^2} \times(3 x)^2\left[\frac{1-\cos x}{\sin x}\right]^2} \\ & =\lim _{x \rightarrow 0} \frac{2 x \tan x\left(1-1+\tan ^2 x\right) \times \sin ^2 x}{\left(1-\tan ^2 x\right) \frac{1-\cos 3 x}{(3 x)^2}} \\ & \times 9 x^2 \frac{(1-\cos x)^2}{x^2} \times x^4 \\ & =\lim _{x \rightarrow 0} \frac{2}{9} \times \frac{\tan x}{x} \times \frac{\tan ^2 x}{\frac{1}{2}} \times \frac{\sin ^2 x}{x^4 \times\left(\frac{1}{2}\right)^2} \\ & =\lim _{x \rightarrow 0} \frac{2}{9} \times 2 \times 4=\frac{16}{9} \end{aligned} \end{aligned} $

Which is continuous and strictly increasing

(B) $f(x)=\sqrt{|x|}=\left\{\begin{aligned} \sqrt{x}, & x \geq 0 \\ \sqrt{-x}, & x<0\end{aligned}\right. f^{\prime}(x)=\frac{1}{2 \sqrt{x}}( \pm 1)$ Which is not defined at $x=0$, this is continuous but not differentiable at $x=0$

(C) $f(x)=x+[x]=\left\{\begin{array}{cc}x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ x+1, & 1 \leq x<2\end{array}\right.$

This is strictly increasing but not differentiable at integer.

$ \begin{aligned}(D) f(x) & =|x-1|+|x|+|x+1| \\ & =\left\{\begin{array}{cc} -3 x, & -1 \leq x<0 \\ 2-x, & 0 \leq x<1 \\ 3 x, & 1 \leq x<\infty \end{array}\right. \end{aligned} $

Clearly, it is not differentiable at $x=-1$, 0,1 but differentiable in $(-1,0) \cup(0,1)$

A-I, B-II, C-V, D-IV

(I) $f(x)= \begin{cases}\frac{1}{2}-x, & x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2 & x \geq \frac{1}{2}\end{cases}$

$f(x)$ is not differentiable at $x=\frac{1}{2}$

LMVT not applicable

(II) $f(x)=|3 x-1|$

$f(x)$ is not differentiable at $x=\frac{1}{3}$

$\therefore$ LMVT is not applicable.

(III) $f(x)=x|x|$ it is continuous and differentiable in $(0,1)$

= LMVT applicable.

(V) $f(x)=|x|$ it is continuous and differentiable in $(0,1)$

$\therefore$ LMVT is applicable.

$ \text { } \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{n+r}{n^2+(r)^2} \\ & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{1+\frac{r}{n}}{1+\left(\frac{r}{n}\right)^2} \\ & =\int_0^2 \frac{1+x}{1+(x)^2} d x \\ & =\int_0^2 \frac{1}{1+x^2} d x+\int_0^2 \frac{x}{1+x^2} d x \\ & =\left(\tan ^{-1} x\right)_0^2+\left(\frac{1}{2} \ln \left(1+x^2\right)\right)_0^2 \\ & =\tan ^{-1} 2+\frac{1}{2} \ln 5 \end{aligned} $

$ \begin{aligned} & \text { } \because \lim _{x \rightarrow 1} f(x) \text { exists } \\ & \therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned} $

$ \begin{aligned} &Now, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ & =\lim _{h \rightarrow 0}\left(1+\frac{2(1-h)}{a}\right)=1+\frac{2}{a} \end{aligned} $

$ \begin{array}{ll} \text { and } & \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ & =\lim _{h \rightarrow 0} a(1+h)=a \\ \therefore & 1+\frac{2}{a}=a \\ \Rightarrow & a+2=a^2 \\ \Rightarrow & a^2-a-2=0 \\ \Rightarrow & a^2-2 a+a-2=0 \\ \Rightarrow & a(a-2)+1(a-2)=0 \\ \Rightarrow & (a-2)(a+1)=0 \\ \Rightarrow & a=-1,2 \end{array} $

$\therefore$ Sum of the cubes of the possible values of $a$ is $-1+8=7$.

At $x=0$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}}\left(\left[x^2\right]-[x]^2\right)=0-1=-1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 0^{+}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 \end{aligned} $

$\therefore$ LHL $\neq$ RHL

Limit does not exists at $x=0$.

At $x=1$

$ \mathrm{LHL}=\lim _{x \rightarrow 1^{-}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 $

$\mathrm{RHL}=\lim _{x \rightarrow 1^{+}}\left(\left[x^2\right]-[x]^2\right)=1-1=0$

As LHL $=$ RHL, so limit exits at $x=1$

At $x=2$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 2^{-}}\left(\left[x^2\right]-[x]^2\right)=3-1=2 \\ & \mathrm{RHL}=\lim _{x \rightarrow 2^{+}}\left(\left[x^2\right]-[x]^2\right)=4-4=0 \end{aligned} $

As LHL $\neq$ RHL, so limit does not exists. Hence, there are two integral points where the limits does not exists.

$ \begin{aligned} &\text { As } f(x) \text { is continuous at } x=0\\ &\begin{aligned} & f(0)=\lim _{x \rightarrow 0} f(x) \\ & K=\lim _{x \rightarrow 0} \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \Rightarrow K=\lim _{x \rightarrow 0} \\ & \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \quad \quad \times \frac{\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}}{\left.\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}\right)} \\ & \quad \quad \times \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-2 a x(\sqrt{a+x}+\sqrt{a-x)}}{2 x\left(\sqrt{x^2-a x+a^2}+\sqrt{x^2+a x+a^2}\right)} \\ & =\frac{-a(2 \sqrt{a})}{2 a}=-\sqrt{a} \\ & \therefore \quad K=-\sqrt{a} \end{aligned} $

$f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \\ b x^3+1, & x > 2\end{array}\right.$

$ f^{\prime}(x)=\left\{\begin{array}{cc} 2 a x+b, & x \leq 1 \\ 3, & 1 < x \leq 2 \\ 3 b x^2, & x > 2 \end{array}\right. $

As, the function is differentiable at $x \in R$,

$ \begin{aligned} & \lim _{x \rightarrow 1} 2 a x+b=3 \text { and } \lim _{x \rightarrow 2} 3 b x^2=3 \\ & \Rightarrow 2 a+b=3 \text { and } 12 b=3 \Rightarrow b=\frac{1}{4} \\ & \quad 2 a=3-\frac{1}{4} \Rightarrow 2 a=\frac{11}{4} \Rightarrow a=\frac{11}{8} \\ & \therefore \quad a-b=\frac{11}{8}-\frac{1}{4}=\frac{9}{8} \end{aligned} $

For option (a), $f(x)=|x|=x$ as $1 \leq x \leq 5$

As $f(x)$ is a polynomial function in its domain, so it is continuous and differentiable.

$ \begin{aligned} & f^{\prime}(x)=1, f(1)=1 \text { and } f(3)=5 \\ & f^{\prime}(x)=\frac{f(5)-f(1)}{5-1}=1 \in[1,5] \end{aligned} $

$\therefore$ Lagrange's mean value theorem is applicable.

For option (b)

$ f(x)=[x]=1 \text { for } x \in[\sqrt{2}, \sqrt{3}] $

As $f(x)$ is a constant function, so Lagranye's mean value theorem is applicable.

For option (c),

The function $f(x)=x^2-1$ is continuous if $x^2-1>0$ i.e. either $x<-1$ or $x>1$.

For $x=\frac{1}{e}, x^2-1>0$, but $\frac{1}{e^2}-1<0$

So, the value is not within the domain Thus, lagrange mean value theorem is not applicable.

For option (d), $f(x)=e^x$ is continuous and differentiable over $[-e, e]$

$ f^{\prime}(x)=e^x \text { and } f(-e)=\frac{1}{e^i} f(l)=e^e{} $

$ \begin{aligned} f^{\prime}(e) & =\frac{f(b)-f(a)}{b-a}=\frac{e^e-\frac{1}{e^e}}{e+e} \\ & =\frac{\left(e^e\right)^2-1}{e^e \times 2 e} \\ & =\frac{e^{e-1}}{2}-\frac{1}{2 e^{e+1}} \in[-e, e] \end{aligned} $

$\therefore$ Lagrange mean value theorem is applicable.

$ \begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0 \end{array}\right. \\ & = \begin{cases}2, & x<0 \\ 0, & x>0 \\ 2, & x=0\end{cases} \\ & \quad\left[\because|x|=\left\{\begin{array}{c} x, \text { if } x>0 \\ -x, \text { if } x<1 \end{array}\right.\right. \end{aligned} $

$\therefore f(x)$ has maximum value 2 and minimum value 0 .

$\mathop {\lim }\limits_{x \to \infty } \frac{[2 x-3]}{x} \text { } $

$ \begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{2 x}{x}-\frac{3}{x}\right) \Rightarrow 2-\lim _{x \rightarrow \infty} \frac{3}{x} \\ & \Rightarrow \quad 2-0 \Rightarrow 2 \end{aligned} $

$\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 3 x}{\cos 4 x-\cos 5 x} \quad\left(\frac{0}{0}\right.$ form $)$

By L' Hospital rule,

$ =\mathop {\lim }\limits_{x \to 0} \frac{-2 \sin 2 x+3 \sin 3 x}{-4 \sin 4 x+5 \sin 5 x} \quad\left(\frac{0}{0} \text { form }\right) $

Again by L' Hospital rule

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-4 \cos 2 x+9 \cos 3 x}{-16 \cos 4 x+25 \cos 5 x} \\ & =\frac{-4 \times 1+9 \times 1}{-16 \times 1+25 \times 1}=\frac{5}{9} \end{aligned} $

Given that

$ f(x)=\left\{\begin{array}{cc} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6} & \text { for } x \neq 3 \\ l & \text { for } x=3 \end{array}\right. $

is continuous at $x=3$

$ \begin{aligned} & \Rightarrow \text { RHL }=\text { LHL }=\text { Value of function } \\ & \therefore \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}=\text { Value of } \end{aligned} $

function at $x=3$

$ \begin{aligned} & \Rightarrow \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}=l \\ & \Rightarrow \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{(3 x+2)(x-3)}=l \end{aligned} $

For limit to be exist, numerator also have a factor $(x-3)$ so

$ 2 x^2+(k+2) x+9=(x-3)(2 x+a) $

Expand and equate the coefficients

$ \begin{aligned} & k+2=a-6 \\ & 9=-3 a \Rightarrow a=-3 \Rightarrow k=-11 \end{aligned} $

So, $2 x^2+(k+2) x+9=(x-3)(2 x-3)$

Then, $\lim \limits_{x \rightarrow 3^{+}} \frac{(x-3)(2 x-3)}{(3 x+2)(x-3)}=1$

$ \Rightarrow \quad \frac{6-3}{9+2}=l \Rightarrow l=\frac{3}{11} $

So, $l-k=\frac{3}{11}+11=\frac{124}{11}$

$\mathop {\lim }\limits_{x \to o} \left[ {{1 \over x} - {1 \over {{e^x} - 1}}} \right] = \mathop {\lim }\limits_{x \to o} {{{e^x} - 1 - x} \over {x({e^x} - 1)}}$

$ \text { Using L'Hospital's rule } $

$=\mathop {\lim }\limits_{x \to o} {{{e^x} - 1} \over {{e^x} - 1 + x{e^x}}}$

$ \text { Again, using L'Hospital rule, we get } $

$ = \mathop {\lim }\limits_{x \to o} {{{e^x}} \over {{e^x} - {e^x} + x{e^x}}} = {{{e^o}} \over {2{e^o} + 0}} = {1 \over 2}.$

$f(x)= \begin{cases}0, & x=0 \\ 2-x, & \text { for } 0 < x < 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 < x < 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{cases}$

continuity at $x=1$

As $x$ approaches 1 from the left

$\mathop {\lim }\limits_{x \to {1^ + }} \ f {(x) = } {1 \over 2} - 1 = - {1 \over 2}$

The value of $f(x)$ at $x=1$

$ f(1)=2 $

LHL $\neq$ RHL

Therefore, $f(x)$ is not continuous at $x=1$

Continuity at $x=2$

As $x$ approaches 2 from the left (1 < x < 2)

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = {1 \over 2} - 2 = - {3 \over 2}$

The value of $f(x)$ at $x=2$

$ f(x)=-\frac{3}{2} $

The LHL and actual value of the function at $x=2$ are equal to RHL

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2) = - {3 \over 2}$

$ \text { Therefore, } f(x) \text { is continuous at } x=2 $

$f(x)=\left(\frac{1+x}{1-x}\right)^{1 / x}$

We will calculate the limit of this expression as $x$ approaches 0 .

Let $L=\left(\frac{1+x}{1-x}\right)^{1 / x}$

$L=\lim _{x \rightarrow 0} \frac{1}{x} \ln \left(\frac{1+x}{1-x}\right)^{1 / x} \ldots(\mathrm{i})($ natural $\log$ )

Let $y=\log \frac{1+x}{1-x}$

$ =\log (1+x)-\log (1-x) $

Using Taylor's expansions for $\log (1+x)$ and $\log (1-x)$ around $x=0$

$ \log (1+x) \approx x $

$ \log (1-x) \approx-x $

Then, $y=x-(-x)$

$ y=2 x \Rightarrow \log \left(\frac{1+x}{1-x}\right)=2 x \ldots \text { Eq. } $

From Eq. (i), we get

$ \begin{aligned} \log L=\lim _{x \rightarrow 0} \frac{1}{x} & \ln \left(\frac{1+x}{1-x}\right) \\ & =\lim _{x \rightarrow 0} \frac{1}{x} \times 2 x \text { using } \\ \log L & =\lim _{x \rightarrow 0} 2 \\ \log L & =2 \\ f(x) & =L=e^2 \end{aligned} $

$\because f(x)$ is continuous at, $x=0, f(0)=e^2$

The function $ f(x) = |x - 24| $ is being evaluated for its continuity and differentiability on the interval $[0, 25]$.

Continuity

The function $ f(x) = |x - 24| $ is continuous over the interval $[0, 25]$. The absolute value function is continuous across its domain, including the point $ x = 24 $ within this interval.

Differentiability

The function $ f(x) = |x - 24| $ is differentiable on the interval $(0, 25)$ except at the point $ x = 24 $. At $ x = 24 $, the left-hand derivative is $ f'(24) = -1 $ and the right-hand derivative is $ f'(24) = 1 $. Since the left-hand and right-hand derivatives at $ x = 24 $ are not equal, the function is not differentiable at this point.

For the subintervals $ 0 < x < 24 $ and $ 24 < x \leq 25 $, the derivative of $ f(x) $ is well defined.

Conclusion

Thus, the function $ f(x) = |x - 24| $ is continuous on $[0, 25]$ but not differentiable on $[0, 25]$, specifically due to the point $ x = 24 $.

$ \begin{aligned} &\text { We can observe that, }\\ &\begin{aligned} & T_n=\frac{1}{\sqrt{n^2-r^2}} \\ & \Rightarrow \lim _{n \rightarrow \infty} \sum_{n=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}=\lim _{n \rightarrow \infty} \sum_{n=0}^{n-1}{1 \over {n\sqrt {1 - {{\left( {{r \over n}} \right)}^1}} }} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } \frac{r}{n}=x \\ & \frac{1}{n}=d x \\ & \text { limit } r=0 \Rightarrow x=0 \\ & r=n-1 \\ & \Rightarrow \quad x=1-\frac{1}{n} \\ & x=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because n=\infty)\\ & \Rightarrow \int_0^1 \frac{d x}{\sqrt{1-x^2}} \\ & =\left(\sin ^{-1}\right)_0^1 \\ & =\sin ^{-1}(\mathrm{l})-\sin ^{-1}(0) \\ & =\frac{\pi}{2}-0=\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \Rightarrow \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} \\ & {\left[\because \sin ^2 x+\cos ^2 x=1\right] } \end{aligned} \end{aligned} $

$ \begin{aligned} & =\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} \frac{\sin \left(\pi \sin ^2 x\right)}{\left(\pi \sin ^2 x\right)} \times \frac{\left(\pi \sin ^2 x\right)}{\left(x^2\right)} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\left(\pi \sin ^2 x\right)} \cdot \lim _{x \rightarrow 0} \frac{\pi \sin ^2 x}{x^2} \\ & =\pi \quad\left\{\because \lim _{x \rightarrow 0} \frac{\sin \theta}{\theta}=1\right\} \end{aligned} $