Question

Calculate each of the limits in Exercises $21-48$. Some of these limits are made easier by L’Hopital’s rule, and some are not.

$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{{\sin }^{-1}x}$.

Short answer

The exact value of the limit$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{{\sin }^{-1}x}$is,$1.$

Step-by-step solution

Step 1. Given information

$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{{\sin }^{-1}x}$.

Step 2. The limit using L'Hopital's rule is given below:

$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{{\sin }^{-1}x}=\underset{x\to 0}{\lim }\frac{\frac{1}{1+x^2}}{\frac{1}{\sqrt{1-x^2}}}$[ L'Hopital's rule]

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{\sqrt{1-x^2}}{1+x^2} \\ =\frac{\sqrt{1-0}}{1+0} \\ =1 \end{gathered}$$

Therefore, the exact value is,$1.$