Question

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

Answer 1

Answer: 3

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