Question

If $\displaystyle\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^3}=L$, then the value of $(4 L+1)$ is equal to

Answer 1

Let $L=\displaystyle\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^3}$
It is $\frac{0}{0}$ form. So, using L'Hospital's rule, we get
$L=\displaystyle\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}-\frac{1}{1+x^2}}}{3 x^2}$
$=\frac{1}{3} \displaystyle\lim _{x \rightarrow 0} \frac{1}{x^2} \frac{1+x^2-\sqrt{1-x^2}}{1+x^2 \sqrt{1-x^2}}$
$=\frac{1}{3} \displaystyle\lim _{x \rightarrow 0} \frac{1}{x^2} \frac{1+x^2-1-x^2}{1+x^2 \sqrt{1-x^2}} \cdot \frac{1}{1+x^2+\sqrt{1-x^2}}$
$=\frac{1}{3}\displaystyle \lim _{x \rightarrow 0} \frac{\left(3+x^2\right)}{\left(1+x^2\right) \sqrt{1-x^2}} \frac{1}{\left[\left(1+x^2\right)+\sqrt{1-x^2}\right]}$
$L=\frac{1}{3} \frac{3}{2}=\frac{1}{2}$
$\Rightarrow 4 L=2 \Rightarrow 4 L+1=3$