# Question #597c8

Jan 8, 2017

The limit is 1.

#### Explanation:

${\lim}_{x \rightarrow \infty} \frac{x}{x + \sin x} = {\lim}_{x \rightarrow \infty} \frac{1}{1 + \left(\sin \frac{x}{x}\right)}$

Note that ${\lim}_{x \rightarrow \infty} \sin \frac{x}{x} = 0$ (by the squeesze theorem).

So

${\lim}_{x \rightarrow \infty} \frac{x}{x + \sin x} = {\lim}_{x \rightarrow \infty} \frac{1}{1 + \left(\sin \frac{x}{x}\right)}$

$= \frac{1}{1 + 0} = 1$

Note on l'Hospital's Rule

Although the initial form is $\frac{\infty}{\infty}$, when we try to apply l'Hospita'l, we get

${\lim}_{x \rightarrow \infty} \frac{1}{1 + \cos x}$.

This limit does not exist, but it is not infinite. Therefore, l'Hospital gives us no information about the limit.