Question

$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^2}$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\infty$

Answer 1

The correct option is C $0$

${lim}_{x\to 0}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^2}$
$={lim}_{x\to 0}\frac{1+x^2-\sqrt{1-x^2}}{2\left(\sqrt{1-x^2}\left(x(x^2+1)\right)\right)}$ (Applying L'Hospital's rule)
$={lim}_{x\to 0}\frac{1+x^2-\sqrt{1-x^2}}{2(1-x^2\left(x(x^2+1)\right))}=\frac{1}{2}{lim}_{x\to 0}\frac{1}{\sqrt{1-x^2}(x^2+1)}\frac{1+x^2-\sqrt{1-x^2}}{x}=\frac{1}{2}\frac{1}{1}{lim}_{x\to 0}\frac{1+x^2-\sqrt{1-x^2}}{x}$
$=\frac{1}{2}{lim}_{x\to 0}(2x+\frac{x}{\sqrt{1-x^2}})=0$ (Applying L'Hospital's rule)
Hence, option 'C' is correct.