Q. 48 - Calculus

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.

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

Step-by-Step Solution

Answer

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

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, $1.$