Question

$\underset{x\to 0}{lim}\frac{{\tan }^{-1}x-{\sin }^{-1}x}{{\sin }^{3}x}$

Answer 1

$\begin{array}{c}{lim}_{x\to 0}\frac{{\tan }^{-1}x-{\sin }^{-1}x}{{\sin }^{2}x}\\ {lim}_{x\to 0}\frac{{\tan }^{-1}x-{\tan }^{-1}\frac{x}{\sqrt{1-x^2}}}{x^2\left(\frac{{\sin }^{3}x}{x^3}\right)}\\ {lim}_{x\to 0}\frac{{\tan }^{-1}x-{\tan }^{-1}\frac{x}{\sqrt{1-x^2}}}{x^3}{lim}_{x\to 0}\frac{\sin x}{x}=1\\ {lim}_{x\to 0}\frac{{\tan }^{-1}\left(\frac{x-\frac{x}{\sqrt{1-x^2}}}{1+\frac{x.x}{\sqrt{1-x^2}}}\right)}{x^3}\\ {lim}_{x\to 0}\frac{{\tan }^{-1}\left(\frac{x\sqrt{1-x^2}-x}{x^2+\sqrt{1-x^2}}\right)}{x^3\left(\frac{x\sqrt{1-x^2}-x}{x^2+\sqrt{1-x^2}}\right)}.\frac{x\sqrt{1-x^2}-x}{x^2+\sqrt{1-x^2}}\\ {lim}_{x\to 0}\frac{x\left(\sqrt{1-x^2}-1\right)}{x^3\left(x^2+\sqrt{1-x^2}\right)}\\ {lim}_{x\to 0}\frac{\left(\sqrt{1-x^2}-1\right)}{x^2\left(x^2+\sqrt{1-x^2}\right)}\times \frac{\sqrt{1-x^2}+1}{\sqrt{1-x^2}+1}\\ {lim}_{x\to 0}\frac{-x^2}{x^2\left(x^2+\sqrt{1-x^2}\right)}\left[\frac{1}{\sqrt{1-x^2}+1}\right]\\ =\frac{-1}{2}\end{array}$

Hence, this is the answer.