# Question #51a7e

May 3, 2015

No the limiti is $0$, because when $x \rightarrow \infty$, $\frac{1}{x} \rightarrow 0$ and so $\sin 0 = 0$.

These are limits they don't exist:

${\lim}_{x \rightarrow + \infty} \sin x$

or

${\lim}_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$.

($\sin \infty$ does not exist).

May 3, 2015

If someone told you the limit does not exist for that reason, they've probably confused this question

${\lim}_{x \rightarrow \infty} \sin \left(\frac{1}{x}\right)$ which is $0$

With this one

${\lim}_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$ which down not exist because the values cover $\left[- 1 , 1\right]$ over shorter and shorter intervals as $x \rightarrow 0$

May 3, 2015

Actually, that WOULD be correct if you were finding the limit of $\sin \left(x\right)$. As $x$ approaches infinity, $\sin \left(\frac{1}{x}\right)$ just becomes $\sin \left(0\right)$, which is $0$. graph{sin(1/x) [-9.775, 10.225, -4.78, 5.22]}