Question

L’Hˆopital’s Rule limit calculations: Calculate each of the

limits that follow. Some of these limits are easier to calculate

by using L’Hopital’s rule, and some are not.

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

Short answer

Limit is$\infty$and it is easy to find using L'Hospital rule

Step-by-step solution

Step 1. Given information

An expression is given as$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{1-\cos x}$

Step 2. Evaluating limit

Use properties of limits,

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{1-\cos x}=\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x}{1-\cos x} \\ =\underset{x\to 0}{\lim }\frac{\frac{1}{1+x^2}}{0-(-\sin x)} \\ =\underset{x\to 0}{\lim }\frac{1}{\sin x}\frac{1}{1+x^2} \\ =\frac{1}{\sin 0}\frac{1}{1+0^2} \\ =\frac{1}{0}\times 1 \\ =\infty \end{gathered}$$