Question

If $\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}$ equals $L$, then the value of $(4L+1)$ is

Answer 1

Answer: 3

Let $L=\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}}{3x^2}(L^{\prime }$ Hospital rule $)$
$=\frac{1}{3}\underset{x\to 0}{\lim }\frac{1}{x^2}\left[\frac{1+x^2-\sqrt{1-x^2}}{\left(1+x^2\right)\sqrt{1-x^2}}\right]$
$=\frac{1}{3}\underset{x\to 0}{\lim }\frac{1}{x^2}\left[\frac{{\left(1+x^2\right)}^2-\left(1-x^2\right)}{\left(1+x^2\right)\sqrt{1-x^2}}\cdot \frac{1}{\left(1+x^2\right)+\sqrt{1-x^2}}\right]$
$=\frac{1}{3}\underset{x\to 0}{\lim }\frac{\left(3+x^2\right)}{\left(1+x^2\right)\sqrt{1-x^2}}\frac{1}{\left[\left(1+x^2\right)+\sqrt{1-x^2}\right]}$
$=\frac{1}{3}\left(\frac{3}{2}\right)=\frac{1}{2}$