Limits, Continuity and Differentiability

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

$ \begin{aligned} & =4 \lim _{x \rightarrow 0} \frac{1}{1-\tan ^2 x} \times \lim _{x \rightarrow 0} \frac{\tan ^3 x}{x^3} \times \lim _{x \rightarrow 0} \frac{x^2 / 4}{\frac{\sin ^2 x}{4}} \times \frac{1}{(-1)} \lim _{x \rightarrow 0} \frac{x}{2^x-1} \\ & =4\left[1 \times 1 \times 1 \times \frac{1}{\log 2}\right]=\frac{4}{\log 2} \end{aligned} $

$ \text { } \begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi\left(1+\tan ^2 x\right)^2\right]}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi+\pi \tan ^4 x+2 \pi \tan ^2 x\right)}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \tan ^4 x+2 \pi \tan ^2 x\right)}{\pi^2 x^4} \\ & {[\because \tan (\pi+\theta)=\tan \theta] } \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]}{\pi^2 x^4} \times \frac{\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]^2}{\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]^2} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\pi^2 \tan ^4 x\left(\tan ^2 x+2\right)^2}{\pi^2 x^4}\times\left\{\frac{\tan \left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]}{\pi \tan ^2 x\left(\tan ^2 x+2\right)}\right\}^2 \\ & \Rightarrow(1)^4 \times(0+2)^2 \times(1)^2=4 \\ & \quad\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right] \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{r} \lim _{x \rightarrow 0}\left(\frac { 4 ! } { x ^ { 8 } } \left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}\right.\right. \left.\left.+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right) \\ =\lim _{x \rightarrow 0}\left(\frac { 4 ! } { x ^ { 8 } } \left[1\left(1-\cos \frac{x^2}{3}\right)\right.\right. \left.\left.-\cos \frac{x^2}{4}\left(1-\cos \frac{x^2}{3}\right)\right]\right) \end{array} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0}\left[\frac{4!}{x^8}\left(1-\cos \frac{x^2}{3}\right)\left(1-\cos \frac{x^2}{4}\right)\right] \\ & =4!\lim _{x \rightarrow 0}\left[2 \sin ^2 \frac{x^2}{6} \times 2 \sin ^2 \frac{x^2}{8}\right] \times \frac{1}{x^8} \\ & =\frac{4 \times 4!}{64 \times 36} \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{x^2}{6}}{\frac{x^4}{36}} \times \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{x^2}{8}}{\frac{x^4}{64}} \\ & =\frac{4 \times 24}{64 \times 36}=\frac{1}{24} \end{aligned} $

According to the question,

$ \begin{aligned} & m=a_{11}+a_{22}+\ldots .+a_{n n} \\ \Rightarrow \quad & m=k+k^2+\ldots .+k^n \end{aligned} $

Again, $\lim _{k \rightarrow 1} \frac{n-m}{1-k}=171$

$ \Rightarrow \lim _{k \rightarrow 1} \frac{m-n}{k-1}=171 $

$ \begin{gathered} \Rightarrow \lim _{k \rightarrow 1} \frac{\left(k+k^2+k^3+\ldots+k^n\right)-(1+1+\ldots \text { to } n \text { times })}{(k-1)}=171 \\ \Rightarrow \lim _{k \rightarrow 1}\left[\frac{(k-1)}{(k-1)}+\frac{\left(k^2-1\right)}{(k-1)}+\frac{\left(k^3-1\right)}{(k-1)}\right. \\ \left.+\ldots+\frac{\left(k^n-1\right)}{(k-1)}\right]=171 \end{gathered} $

$ \begin{aligned} & \Rightarrow 1+2+3+\ldots+n=171 \\ & \quad\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] \\ & \Rightarrow \frac{n(n+1)}{2}=171 \\ & \Rightarrow \quad n=18 \end{aligned} $

$ \begin{aligned} & \text { 66. (c) } \lim _{x \rightarrow \infty} x^3\left[\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2} x\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\begin{array}{l} \left(\sqrt{x^2+\sqrt{x^4+1}}\right. \\ \frac{-\sqrt{2} x)\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)} \end{array}\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\frac{x^2+\sqrt{x^4+1}-2 x^2}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)}\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\frac{\left(\sqrt{x^4+1}-x^2\right)\left(\sqrt{x^4+1}+x^2\right)}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)\left(\sqrt{x^4+1}+x^2\right)}\right] \end{aligned} $

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

At $x=-3$

$ \begin{aligned} \mathbf{L H L} & =\lim _{h \rightarrow 0} f(-3-h) \\ & =\lim _{h \rightarrow 0}(3-(-3-h))=6 \end{aligned} $

RHL $\lim _{h \rightarrow 0} f(-3+h)=6 \Rightarrow f(-3)=6$

$\Rightarrow f(x)$ is continuous at $x=-3$

At $x=3$

$ \begin{aligned} \mathbf{L H L} & =\lim _{h \rightarrow 0} f(3-h)=6 \\ \mathbf{R H L} & =\lim _{h \rightarrow 0} f(3+h) \\ & =\lim _{h \rightarrow 0}(3+(3+h))=6 \\ f(3) & =6 \end{aligned} $

$\Rightarrow f(x)$ is continuous at $x=3$

$\therefore f(x)$ is continuous $\forall x \in R$

$ \therefore \alpha=0 $

Again, $f(x)=\left\{\begin{array}{cc}3-x, & x<-3 \\ 6, & -3 \leq x \leq 3 \\ 3+x, & x>3\end{array}\right.$

$ f^{\prime}(x)=\left\{\begin{array}{cc} -1, & x<-3 \\ 0, & -33 \end{array}\right. $

$ \begin{aligned} \because \quad f^{\prime}\left(-3^{-}\right) & =-1 \text { and } f^{\prime}\left(-3^{+}\right)=0 \\ f^{\prime}\left(-3^{-}\right) & \neq f^{\prime}\left(-3^{+}\right) \end{aligned} $

$\Rightarrow f(x)$ is not differentiable at $x=-3$ and $f^{\prime}\left(3^{-}\right)=0$ and $f^{\prime}\left(3^{+}\right)=1 \neq f^{\prime}\left(3^{+}\right)$

$\Rightarrow f(x)$ is not differentiable at $x=3$

$ \therefore \beta=2 $

Hence, $\alpha+\beta=0+2=2$.

$ \begin{aligned} & \text { Given, } \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3} \\ & \lim _{x \rightarrow 3^{-}} \frac{(x-2)\left(x^2-x-6\right)}{(x-1)\left(2 x^2-5 x-3\right)} \\ & =\lim _{x \rightarrow 3^{-}} \frac{(x-2)(x-3)(x+2)}{(x-1)(2 x+1)(x-3)}=\frac{1 \times 5}{2 \times 7}=\frac{5}{14} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\left(2^x\right)^2-2 \cdot 2^x+1+1-\cos 2 x}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2+2 \sin ^2 x}{x^2} \\ & =\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^2+2\left(\frac{\sin x}{x}\right)^2=(\ln 2)^2+2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &f(x)=\left\{\begin{array}{ll} \frac{x^2-16}{x-4} ; & x>4 \\ 2 x ; & x \leq 4 \end{array}= \begin{cases}x+4 ; & x>4 \\ 2 x ; & x \leq 4\end{cases}\right. \end{aligned} $

$ \begin{aligned} f^{\prime}\left(4^{-}\right) & =\lim _{h \rightarrow 0} \frac{f(4-h)-f(4)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{2(4-h)-8}{-h}=+2 \\ f^{\prime}\left(4^{+}\right) & =\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(4+h+4)-8}{h}=1 \\ & f^{\prime}\left(4^{-}\right)+f^{\prime}\left(4^{+}\right)=3 \end{aligned} $

$ \begin{array}{r} \text {Given that, } \mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \frac{(1-\cos x)^2}{x^4} \times \frac{x^4}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \mathop {\lim }\limits_{x \to 0} \left(\frac{1-\cos x}{x^2}\right)^2 \\ \times \mathop {\lim }\limits_{x \to 0} \left(\frac{x}{\sin x}\right)^4 \end{array} $

$ \begin{aligned} & \text { As } x \rightarrow 0 \\ & (1-\cos x) \rightarrow 0 \\ & =\lim _{(1-\cos x) \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \\ & \quad \times \lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{x^2} \times \lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^4 \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times(1)^2 \\ & \left\{\because \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2} \text { and } \lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right\} \\ & =\frac{1}{8} \end{aligned} $

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x}{|x|+2 x^2}, & x \neq 0 \\ k, & x=0 \end{array}\right. \\ & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{-h}{|-h|+2(-h)^2}=\frac{-h}{h+2 h^2}=\frac{-h}{h(1+2 h)} \\ & =\lim _{h \rightarrow 0} \frac{-1}{1+2 h}=\frac{-1}{1+2 \times 0}=-1 \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ & \quad=\lim _{h \rightarrow 0} \frac{h}{|h|+2 h^2}=\frac{h}{h(1+2 h)} \\ & \quad=\lim _{h \rightarrow 0} \frac{1}{1+2 h}=\frac{1}{1+2 \times 0}=1 \end{aligned} $

Also, $f(0)=k$

∵ LHL $\neq$ RHL, therefore given function is discontinuous for all values of $k$.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\ & {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\ & =\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{-\cos x}{1}\left[\text { By } L^{\prime} \text { Hospital's rule }\right] \\ & =-\cos k \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} f(x) & = \begin{cases}1+x, & 0 \leq x \leq 2 \\ 3-x, & 2< x \leq 3\end{cases} \\ f(f(x)) & = \begin{cases}1+f(x), & 0 \leq f(x) \leq 2 \\ 3-f(x), & 2< f(x) \leq 3\end{cases} \\ f f(x) & = \begin{cases}1+(1+x), & 0 \leq x \leq 1 \\ 3-(1+x), & 1< x \leq 2 \\ 1+(3-x), & 2< x \leq 3\end{cases} \\ f \circ f(x) & = \begin{cases}2+x, & 0 \leq x \leq 1 \\ 2-x, & 1< x \leq 2 \\ 4-x, & 2< x \leq 3\end{cases} \end{aligned} $

$\therefore f \circ f(x)$ is discontinuous at $x=1$, and

$ \begin{aligned} & x=2 \\ & \therefore \quad a=1, b=2 \\ & \quad 2 a+3 b=2(1)+3(2)=2+6=8 \end{aligned} $

We have,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right.}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi+\pi x+x^2\right)}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(\pi x+x^2\right)}{x^2} \\ & {[\because \cos (2 \pi+\theta)=\cos \theta]} \end{aligned} $

Applying 'L' Hospital rule

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

Again apply 'L' Hospital rule

$ \begin{aligned} & =\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\cos \left(\pi x+x^2\right)(2 x+\pi)}{1} \\ & =\frac{\pi}{2} \times \pi=\frac{\pi^2}{2} \end{aligned} $

Given,

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

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

$ \therefore \quad(\mathrm{LHL})_{X=0}=(\mathrm{RHL})_{X=0}=f(0) $

Now, (LHL)

$ \begin{aligned} x=0 & =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 4(0-h)}{(0-h)^2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{h^2}=\lim _{h \rightarrow 0} \frac{2 \sin ^2 2 h}{h^2} \\ & =2 \lim _{h \rightarrow 0}\left(\frac{\sin 2 h}{2 h}\right)^2 \times 4 \\ & =2 \times(1)^2 \times 4=8 \end{aligned}\\ &\text { From Eq. (i), we have }\\ &\begin{aligned} & (\mathrm{LHL})_x=0=f(0) \Rightarrow 8=a \\ \Rightarrow \quad & a=8 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{(1+x) \log (1+x)-x}{x^2}=k \\ \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)+1-1}{2 x}=k \\ \Rightarrow & \quad \text { [using L' Hospital Rule] } \\ \Rightarrow & \frac{1}{2} \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=k \\ \Rightarrow & \frac{1}{2} \times 1=k \quad\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=1\right] \\ \Rightarrow & k=\frac{1}{2} \quad \therefore \quad 12 k=12 \times \frac{1}{2}=6 \end{array} $

$ \begin{aligned} & \text { We have, } \\ & f(x) \text { is continuous at } x=1 \\ & \therefore \quad k=\lim _{x \rightarrow 1} \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9 x^{3 / 2}-9 x-\sqrt{x}+1}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9\left(\frac{3}{2} x^{1 / 2}\right)-9-\frac{1}{2 \sqrt{x}}}{6 x+2} \\ & \Rightarrow \quad k=\frac{\frac{27}{2}-9-\frac{1}{2}}{6+2}=\frac{13-9}{8}=\frac{4}{8}=\frac{1}{2} \end{aligned} $

Let $f(x)=\frac{2 x-1}{3 x-4},[1,2]$

Since, $f(x)$ is not defined at

$ x=\frac{4}{3} \in[1,2] $

So, Lagrange's theorem is not applicable on $f(x)$ on $[1,2]$