Limits, Continuity and Differentiability

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)\left[1-\cos \left(\frac{\pi}{2}-x\right)\right]}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right) \cdot 2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\left(\frac{\pi}{2}-x\right) \cdot 16\left(\frac{\pi}{4}-\frac{x}{2}\right)^2} \\ & =\frac{2}{32} \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)} \times \lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}{\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}^2=\frac{2}{32}=\frac{1}{16} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} h e^a\left(\frac{e^{h h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} e^a\left\{\frac{e^{b-a}-1}{\left(\frac{e^h-1}{h}\right)}\right\}=e^a\left(e^{b-a}-1\right)=e^b-e^a \end{aligned} $

When $x<2|x-2|=-(x-2)=2-x$.

Then for $x<2$ :

$ \begin{aligned} g(x) & =\frac{x^2-4}{x^2-2(2-x)-4}=\frac{x^2-4}{x^2+2 x-8} \\ & =\frac{(x-2)(x+2)}{(x-2)(x+4)}=\frac{x+2}{x+4} \end{aligned} $

The limit from the left is

$ \mathop {\lim }\limits_{x \to {2^ - }} g(x)=\mathop {\lim }\limits_{x \to {2^ - }} \frac{x+2}{x+4}=\frac{2+2}{2+4}=\frac{4}{6}=\frac{2}{3} $

When $x>2|x-2|=x-2$.

Then, for $x>2$ :

$ \begin{aligned} g(x) & =\frac{x^2-4}{x^2-2(x-2)-4}=\frac{x^2-4}{x^2-2 x} \\ & =\frac{(x-2)(x+2)}{x(x-2)}=\frac{x+2}{x} \end{aligned} $

The limit from the right is:

$ \mathop {\lim }\limits_{x \to {2^ + }} g(x)=\mathop {\lim }\limits_{x \to {2^ + }} \frac{x+2}{x}=\frac{2+2}{2}=\frac{4}{2}=2 $

Since, $\mathop {\lim }\limits_{x \to {2^ - }} g(x)=\frac{2}{3}$ and $\mathop {\lim }\limits_{x \to {2^ + }} g(x)=2$, the limit $\mathop {\lim }\limits_{x \to {2}} g(x)$ does not exist because the left and right limits are not equal.

Since, the limit does not exists at $x=2, g(x)$ is not continuous at $x=2$

We have, $\mathop {\lim }\limits_{x \to \infty }\left\{\frac{x^2-1}{x+1}-a x-b\right\}=2$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{x^2(1-a)-x(a+b)-1-b}{x+1}=2 $

Since, the limit of the given expression is a finite non-zero number, therefore numerator and denominator are of the same degree.

$ \begin{aligned} \therefore & & 1-a & =0 \\ \Rightarrow & & a & =1 \end{aligned} $

$ f(x)=\left\{\begin{array}{cc}\frac{x^{2}-1}{x^{2}-2 x+1} ; & x > 1 \\ \frac{1}{2} ; & x=1 \\ \frac{x^{2}-1}{x^{2}+2 x-3} ; & x < 1\end{array}\right. $

$ \begin{array}{ll} \because & x^{2}+3 x-x-3 \\ & =x(x+3)-1(x+3) \end{array} $

Lets Find

LHL, $ \lim \limits_{x \rightarrow 1^{-}} \frac{x^{2}-1}{x^{2}-2 x+1} $ and RHL, $ \lim \limits_{x \rightarrow 1^{+}} \frac{x^{2}-1}{x^{2}+2 x-3} $

$ \begin{array}{l} \Rightarrow \lim \limits_{x \rightarrow 1^{-}} \frac{(x-1)(x+1)}{(x-1)^{2}} \Rightarrow \lim \limits_{x \rightarrow 1^{+}} \frac{(x-1)(x+1)}{(x+3)(x-1)} \\ \Rightarrow \lim \limits_{x \rightarrow 1^{-}} \frac{x+1}{x-1}=\infty \Rightarrow \lim \limits_{x \rightarrow 1^{+}} \frac{x+1}{x+3}=\frac{2}{4}=\frac{1}{2} \end{array} $

$ \therefore \mathrm{LHL} \neq f(1) \Rightarrow f(x) $ is not continuous at $ x=1 $

$\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2} \cos \frac{x^2}{4}\right) \\ & =\lim _{x \rightarrow 0} \frac{8}{x^8}\left(\left(1-\cos \frac{x^2}{2}\right)-\cos \frac{x^2}{4}\left(1-\cos \frac{x^2}{2}\right)\right) \\ & =\lim _{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \frac{x^2}{4}\right) \end{aligned}$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{8}{x^8}\left(2 \sin ^2 \frac{x^2}{4}\right)\left(2 \sin ^2 \frac{x^2}{8}\right) \\ & =\lim _{x \rightarrow 0} \frac{8 \times 2 \times 2}{4^2 \times 8^2}\left(\frac{\sin \frac{x^2}{4}}{\frac{x^2}{4}}\right)^2\left(\frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}\right)^2 \\ & =\frac{1}{32} \times(1)^2 \times(1)^2=\frac{1}{32} \end{aligned}$

Given, $\lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}}$

$ =\lim _\limits{x \rightarrow 0} \frac{3\left[\left(1+\frac{x}{27}\right)^{1 / 3}-1\right]}{9\left[1-\left(1+\frac{x}{27}\right)^{2 / 3}\right]} $

$ =\lim _\limits{x \rightarrow 0} \frac{1}{3}\left[\frac{\left.\left(1+\frac{x}{3 \times 27}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!} x^2+\ldots\right)-1\right)}{1-\left(1+\frac{2 x}{3 \times 27}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2!} x^2+\ldots\right)}\right] $

$ \begin{aligned} & =\lim _\limits{x \rightarrow 0} \frac{1}{3}\left[\frac{x\left(\frac{1}{3 \times 27}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!} x+\ldots\right.}{-x\left(\frac{2}{3 \times 27}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2!} x+\ldots\right)}\right] \\ & =\frac{1}{3}\left[\frac{\frac{1}{3 \times 27}+0+\ldots .}{-\left(\frac{2}{3 \times 27}+0+\ldots\right)}\right] \\ & =\frac{1}{3}\left(\frac{1}{-2}\right)=-\frac{1}{6} . \end{aligned} $

$1 - \cos (1 - \cos x) = 2{\sin ^2}\left( {{{1 - \cos x} \over 2}} \right) = 2{\sin ^2}\left( {{{\sin }^2}{x \over 2}} \right)$

$\therefore$ $\mathop {\lim }\limits_{x \to 0} {{1-\cos (1 - \cos x)} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}\left( {{{\sin }^2}{x \over 2}} \right)} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}\left( {{{\sin }^2}{x \over 2}} \right)} \over {{{\left( {{{\sin }^2}{x \over 2}} \right)}^2}}} \times {{{{\sin }^4}{x \over 2}} \over {{{\left( {{x \over 2}} \right)}^2} \times 16}}$

[$\because$ $\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1$]

$ = {1 \over 8}$

Let $y = {(1 + x)^{1/x}}$

$ \Rightarrow \log y = {1 \over x}\log (1 + x) = {1 \over x}\left[ {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - {{{x^4}} \over 4}\, + \,...} \right]$

$ = 1 - {x \over 2} + {{{x^2}} \over 3} - {{{x^3}} \over 4}\, + \,....$

$y = {e^{\left( {1 - {x \over 2} + {{{x^2}} \over 3}\, + \,...} \right)}} = e{e^{\left( {{{ - x} \over 2} + {{{x^2}} \over 3}\,....} \right)}}$

$ = e\left[ {1 + \left( { - {x \over 2} + {{{x^2}} \over 3}\, - \,...} \right) + {1 \over {2!}}{{\left( { - {x \over 2} + {{{x^3}} \over 3}\, - \,....} \right)}^2} + \,...} \right]$

$y - e + {1 \over 2}ex = e{x^2}\left[ {{1 \over 3} + 0(x) + {1 \over 2}{{\left( { - {1 \over 2} + 0(x)} \right)}^2}\, + \,...} \right]$ [$\because$ 0(x) in terms containing x]

$\mathop {\lim }\limits_{} {{y - e + {1 \over 2}ex} \over {{x^2}}} = e\left[ {{1 \over 3} + {1 \over 8}} \right] = {{11} \over {24}}e$

The required limit

$ = \mathop {\lim }\limits_{x \to 0} {{\sqrt {(1 - \cos {x^2})(1 + \cos {x^2})} } \over {(1 - \cos x)\sqrt {1 + \cos {x^2}} }}$

$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{{\sin {x^2}} \over {{x^2}}}} \right)} \over {2\left( {{{{{\sin }^2}x/2} \over {{x^2}}}} \right)}}.{1 \over {\sqrt {1 + \cos {x^2}} }} = {1 \over {2 \times {1 \over 4} \times \sqrt 2 }} = \sqrt 2 $

$f'(1 - 0) = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{a{{(1 - h)}^2} + 1 - (1 + a)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{a({h^2} - 2h)} \over { - h}} = 2a$

$f'(1 + 0) = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2} + a(1 + h) + b - (1 + a)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2} + 2h + ah + b} \over h} = 2 + a$, if b = 0

Thus, $2a = 2 + a,b = 0$

$ \Rightarrow a = 2,b = 0$

$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left\{ {{1 \over {n + 1}} + {2 \over {n + 2}} + .... + {{3n} \over {n + 3n}}} \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{3n} {{1 \over n}\left( {{r \over {n + r}}} \right) = \int_0^3 {{x \over {1 + x}}dx} } $

$ = \int_0^3 {\left( {1 - {1 \over {1 + x}}} \right)dx} $

$ = [x - \log (1 + x)]_0^3$

$ = 3 - \log 4 = 3 - 2\log 2$