Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}\left(\frac{0}{0}form\right)$

$=\underset{x\to 0}{lim}\frac{\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}}{3x^2}$ (L' Hospital's rule)

$=\frac{1}{3}\underset{x\to 0}{lim}\frac{1}{x^2}\left[\frac{1+x^2-\sqrt{1-x^2}}{(1+x^2)\sqrt{1-x^2}}\right]$

$=\frac{1}{3}\underset{x\to 0}{lim}\frac{1}{x^2}[\frac{(1+x^2{)}^2-(1-x^2)}{(1+x^2)\sqrt{1-x^2}}.\frac{1}{(1+x^2)+\sqrt{1-x^2}}]$

$=\frac{1}{3}\underset{x\to 0}{lim}\frac{x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}}.\frac{1}{(1+x^2)+\sqrt{1-x^2}}$

$=\frac{1}{3}\left(\frac{3}{2}\right)=\frac{1}{2}$
Hence, option 'D' is correct.