Limit Exercise Solved by two methods without using L'Hospital Rule ,

Exercise:

Show that, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x\arcsin {{x}^{2}}}{x\cos x-\sin x}=-3$

Solution:  Let $\theta =\arcsin {{x}^{2}}\Leftrightarrow {{x}^{2}}=\sin \theta$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{x\arcsin {{x}^{2}}}{x\cos x-\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\theta }{x\left( \cos x-\frac{\sin x}{x} \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\theta }{\cos x-\frac{\sin x}{x}}\times \frac{\sin \theta }{\sin \theta }$

But $\underset{x\to 0}{\mathop{\lim }}\,\frac{\theta }{\sin \theta }=\underset{x\to 0}{\mathop{\lim }}\,\frac{\arcsin {{x}^{2}}}{{{x}^{2}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\arcsin u}{u}\overset{H.R}{\mathop{=}}\,\underset{u\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{1-{{u}^{2}}}}=1$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\theta \sin \theta }{\left( \cos x-\frac{\sin x}{x} \right)\sin \theta }=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\cos x-\frac{\sin x}{x}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\sin \theta }{x\cos x-\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}}{x\cos x-\sin x}$

                                  $=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{{{x}^{3}}}{\cos x}}{x-\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}\sec x}{x-\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}}{x-\tan x}$

Let $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\tan x}{{{x}^{3}}}$

Put $x=3u$ as $x\to 0\,\,,\,\,u\to 0$

So $L=\underset{u\to 0}{\mathop{\lim }}\,\frac{3u-\tan 3u}{27{{u}^{3}}}\Leftrightarrow 27L=\underset{u\to 0}{\mathop{\lim }}\,\frac{3u-\tan 3u}{{{u}^{3}}}$

Recall that $\tan 3u=\frac{{{\tan }^{3}}u-3\tan u}{3{{\tan }^{2}}u-1}$

Hence $27L=\underset{u\to 0}{\mathop{\lim }}\,\frac{3u-\frac{{{\tan }^{3}}u-3\tan u}{3{{\tan }^{2}}u-1}}{{{u}^{3}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\frac{9u{{\tan }^{2}}u-3u-{{\tan }^{3}}u+3\tan u}{3{{\tan }^{2}}u-1}}{{{u}^{3}}}$

              $=\underset{u\to 0}{\mathop{\lim }}\,\frac{9u{{\tan }^{2}}u-3u-{{\tan }^{3}}u+3\tan u}{3{{u}^{3}}{{\tan }^{2}}u-{{u}^{3}}}$

Hence $27L=\underset{u\to 0}{\mathop{\lim }}\,\frac{1}{3{{\tan }^{2}}u-1}\times \left[ \underset{u\to 0}{\mathop{\lim }}\,\frac{3\left( \tan u-u \right)}{{{u}^{3}}}-\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{3}}u}{{{u}^{3}}}+\underset{u\to 0}{\mathop{\lim }}\,\frac{9u{{\tan }^{2}}u}{{{u}^{3}}} \right]$

$\Rightarrow 27L=-3\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u}{{{u}^{3}}}+\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{3}}u}{{{u}^{3}}}-9\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{2}}u}{{{u}^{2}}}$

But $\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u}{{{u}^{3}}}=-\underset{u\to 0}{\mathop{\lim }}\,\frac{u-\tan u}{{{u}^{3}}}=-\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\tan x}{{{x}^{3}}}=-L$

And $\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u}{u}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\frac{\sin u}{\cos u}}{\frac{u}{1}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u\cos u}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u}=1$

Thus $27L=3L+1-9\Leftrightarrow 27L-3L=-8\Leftrightarrow 24L=-8\Leftrightarrow L=-\frac{1}{3}$

Hence $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\tan x}{{{x}^{3}}}=\frac{1}{\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\tan x}{{{x}^{3}}}}=-3$

Thus $\underset{x\to 0}{\mathop{\lim }}\,\frac{x\arcsin {{x}^{2}}}{x\cos x-\sin x}=-3$

2nd method using Series expansion

We know that $\frac{d}{dx}\left( \arcsin {{x}^{2}} \right)=\frac{2x}{\sqrt{1-{{x}^{4}}}}$

And th series of expansion of $\frac{1}{\sqrt{1-{{x}^{2}}}}={{\left( 1-{{x}^{2}} \right)}^{-1/2}}=\sum\limits_{i=0}^{\infty }{\left( \begin{matrix}   -1/2 \\   i \\ \end{matrix} \right)}{{\left( -{{x}^{2}} \right)}^{i}}=1+\frac{{{x}^{2}}}{2}+\frac{3}{8}{{x}^{4}}+\frac{15}{48}{{x}^{6}}+...\,\,\,\,\,,\,\,-1<x\le 1$

$\Rightarrow \frac{2x}{\sqrt{1-{{x}^{4}}}}=2x{{\left( 1-{{x}^{4}} \right)}^{-1/2}}=2x\left( 1+\frac{{{x}^{4}}}{2}+\frac{3{{x}^{8}}}{8}+\frac{15{{x}^{12}}}{48}+.... \right)$    ,  $-1<x\le 1$

Thus $\arcsin {{x}^{2}}=\int_{0}^{x}{\frac{2x}{\sqrt{1-{{x}^{4}}}}dx=\int_{0}^{x}{\left( 2x+{{x}^{5}}+\frac{6{{x}^{9}}}{8}+\frac{30{{x}^{13}}}{48} \right)dx}={{x}^{2}}+\frac{{{x}^{6}}}{6}+\frac{6{{x}^{10}}}{80}+\frac{30{{x}^{14}}}{48\times 14}+..}$

Also we have $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-\frac{{{x}^{7}}}{7!}+.....\,\,\,\And \,\,\,\,\,\cos x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-\frac{{{x}^{6}}}{6!}+...$

Hence $\underset{x\to 0}{\mathop{\lim }}\,\frac{x\arcsin {{x}^{2}}}{x\cos x-\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( {{x}^{2}}+\frac{{{x}^{6}}}{6}+\frac{6{{x}^{10}}}{80}+... \right)}{x\left( 1-\frac{{{x}^{2}}}{2}+\frac{{{x}^{4}}}{24}-\frac{{{x}^{6}}}{6!} \right)-\left( x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-\frac{{{x}^{7}}}{7!}+... \right)}$

                              $=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}\left( 1+\frac{1}{6}{{x}^{3}}+\frac{6}{80}{{x}^{7}}+..... \right)}{x-\frac{{{x}^{3}}}{2}+\frac{{{x}^{5}}}{24}-\frac{{{x}^{7}}}{6!}-x+\frac{{{x}^{3}}}{3!}-\frac{{{x}^{5}}}{5!}+\frac{{{x}^{7}}}{7!}+...}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}\left( 1+\frac{{{x}^{3}}}{6}+\frac{6{{x}^{7}}}{80}+.... \right)}{{{x}^{3}}\left( -\frac{1}{2}+\frac{1}{3!}+\frac{{{x}^{2}}}{24}-\frac{{{x}^{2}}}{5!}-\frac{{{x}^{4}}}{6!}+\frac{{{x}^{4}}}{7!} \right)}=\frac{1}{-\frac{1}{2}+\frac{1}{6}}=\frac{1}{\frac{-6+3}{6}}=\frac{6}{-3}=-3$

*_______________________

The idea of First Method Credit to María Jesús Méndez