Question

Solve ${lim}_{x\to 0}\left(\frac{si{n}^{-1}x-ta{n}^{-1}x}{x^2}\right)$

Answer 1

${lim}_{x\to 0}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^2}\to \frac{0}{0}$ form

so we can try L Hospitals rule

$=\frac{\frac{d}{dx}({\sin }^{-1}x-{\tan }^{-1}x)}{\frac{d}{dx}\left(x^2\right)}$

$=\frac{\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}}{2x}$

$=\frac{\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right)-\frac{d}{dx}\left(\frac{1}{1+x^2}\right)}{\frac{d}{dx}\left(2x\right)}$

$=\frac{\frac{d}{dx}\left({(1-x^2)}^{\frac{-1}{2}}\right)-\left[\frac{-1}{{(1+x^2)}^2.2x}\right]}{2}$

$=\frac{\frac{-1}{2}{(1-x^2)}^{\frac{-1}{2}}-\left[\frac{-1}{{(1+x^2)}^2}2x\right]}{2}$

$=\frac{x\left(\frac{-1}{(1-x^2)\sqrt{1-x^2}}\right)+\frac{2x}{{(1+x^2)}^2}}{2}$

$={lim}_{x\to 0}\frac{\left(\frac{-x}{(1-x^2)\sqrt{1-x^2}}\right)+\frac{2x}{{(1+x^2)}^2}}{2}$

$={lim}_{x\to 0}\frac{\left(\frac{0}{(1-0)\sqrt{1-0}}\right)+\frac{2\times 0}{{(1+0)}^2}}{2}=\frac{0}{2}=0$