Nice Limit Exercise asked in mathematics Stack Exchange page and solved without using Series expansion or L'Hospital Rule

Exercise:

Show that, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{x-\tan x}=-\frac{1}{2}$ without using L’Hospital Rule or Series expansion

Solution: we have \[\frac{x-\sin x}{x-\tan x}=\frac{\tan x-\tan x+x-\sin x}{x-\tan x}=\frac{\left( x-\tan x \right)+\left( \tan x-\sin x \right)}{x-\tan x}\]

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{x-\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( x-\tan x \right)+\left( \tan x-\sin x \right)}{x-\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\tan x}{x-\tan x}+\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{x-\tan x}$

Let $l=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{x-\tan x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x\left( 1-\frac{\sin x}{\tan x} \right)}{\tan x\left( \frac{x}{\tan x}-1 \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\frac{\sin x}{\tan x}}{\frac{x}{\tan x}-1}$

Notice that, $\frac{\sin x}{\tan x}=\frac{\frac{\sin x}{1}}{\frac{\sin x}{\cos x}}=\frac{\sin x\cos x}{\sin x}=\cos x$

So $\frac{1-\cos x}{\frac{x}{\tan x}-1}=\frac{2{{\sin }^{2}}\left( \frac{x}{2} \right)}{\frac{x-\tan x}{\tan x}}=\frac{2\tan x{{\sin }^{2}}\left( \frac{x}{2} \right)}{x-\tan x}=\frac{2\frac{\tan x}{x}\frac{{{\sin }^{2}}\left( x/2 \right)}{{{x}^{2}}}}{\frac{x-\tan x}{{{x}^{3}}}}=\frac{{{l}_{*}}}{{{l}_{\$}}}$

Let ${{l}_{1}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\tan 0}{x-0}=\frac{d}{dx}{{\left( \tan x \right)}_{x=0}}={{\left( {{\sec }^{2}}x \right)}_{x=0}}=1$

Also ${{l}_{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}\left( x/2 \right)}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin \left( x/2 \right)}{2\left( x/2 \right)} \right)}^{2}}=\frac{1}{4}{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \left( x/2 \right)}{x/2} \right)}^{2}}=\frac{1}{4}$

Thus, ${{l}_{*}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\tan x}{x}\times \frac{{{\sin }^{2}}\left( x/2 \right)}{{{x}^{2}}}=2{{l}_{1}}\times {{l}_{2}}=\frac{2}{4}=\frac{1}{2}$

Now we need to find ${{l}_{\$}}=\underset{x\to0}{\mathop{\lim}}\,\frac{x-\tan x}{{{x}^{3}}}=-\underset{x\to0}{\mathop{\lim}}\,\frac{\tan x-x}{{{x}^{3}}}=-\frac{1}{3}$

Let $x=2u\,\,,\,\,as\,\,\,x\to 0,u\to 0$

So ${{l}_{\#}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-x}{{{x}^{3}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan 2u-2u}{8{{u}^{3}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{2u\left( \frac{\tan 2u}{2u}-1 \right)}{8{{u}^{3}}}=\frac{1}{4}\underset{u\to 0}{\mathop{\lim }}\,\frac{\frac{\tan 2u}{2u}-1}{{{u}^{2}}}$

but $\tan 2u=\frac{2\tan u}{1-{{\tan }^{2}}u}$ so $\frac{\frac{2\tan u}{1-{{\tan }^{2}}u}}{2}=\frac{2\tan u}{2u\left( 1-{{\tan }^{2}}u \right)}=\frac{\tan u}{u\left( 1-{{\tan }^{2}}u \right)}$

so $4{{l}_{\#}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\frac{\tan u}{u\left( 1-{{\tan }^{2}}u \right)}-1}{{{u}^{2}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\frac{\tan u-u\left( 1-{{\tan }^{2}}u \right)}{u\left( 1-{{\tan }^{2}}u \right)}}{{{u}^{2}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u+u{{\tan }^{2}}u}{{{u}^{3}}\left( 1-{{\tan }^{2}}u \right)}$

$\Leftrightarrow 4{{l}_{\#}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{1}{1-{{\tan }^{2}}u}\times \underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u+u{{\tan }^{2}}u}{{{u}^{3}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u+u{{\tan }^{2}}u}{{{u}^{3}}}$

$\Leftrightarrow 4{{l}_{\#}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\tan u-u}{{{u}^{3}}}+\underset{u\to 0}{\mathop{\lim }}\,\frac{u{{\tan }^{2}}u}{{{u}^{3}}}={{l}_{\#}}+\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{2}}u}{{{u}^{2}}}={{l}_{\#}}+1\Leftrightarrow {{l}_{\#}}=\frac{1}{3}$

Thus $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{x-\tan x}=1+\frac{1/2}{-1/3}=1-\frac{3}{2}=\frac{2-3}{2}=\frac{-1}{2}$


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The idea of solution Credit to :Amer Khamiseh and شحات جامع