Chapter 14.1, Problem 15AYU

To determine

To find: The limit of $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}$.

Answer to Problem 15AYU

Solution:

$0$

Explanation of Solution

Given:

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

Calculation:

The given limit $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}$ is in $\left(\frac{0}{0}\right)$ form.

So we can apply L’Hospitals’ rule.

$\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=\underset{x\to 0}{\lim }\frac{\left(-\sin x-0\right)}{\left(1\right)}=\underset{x\to 0}{\lim }\left(-\sin x\right)=-\sin \left(0\right)=\text{ 0}$