# How do I find the limit as x approaches infinity of xsin(1/x)?

Sep 26, 2014

By l'Hopital's Rule,

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

Let us look at some details.

${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right)$

by rewriting a bit,

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

by l'Hopital's Rule,

$= {\lim}_{x \to \infty} \frac{\cos \left(\frac{1}{x}\right) \cdot \left(- \frac{1}{x} ^ 2\right)}{- \frac{1}{x} ^ 2}$

by cancelling out $- \frac{1}{x} ^ 2$,

$= {\lim}_{x \to \infty} \cos \left(\frac{1}{x}\right) = \cos \left(0\right) = 1$