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.

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

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

$\underset{x\to 0}{\lim }\frac{\cos x-1}{\sin x}=\underset{x\to 0}{\lim }\frac{-\sin x}{\cos x}$[L'Hopital's rule]

$$\begin{gathered} =-\frac{0}{1} \\ =0 \end{gathered}$$

The exact value is,$0$.