# What is the limit of xsinx as x approaches infinity?

Jul 18, 2016

The limit does not exist. See below.

#### Explanation:

We can determine the result by pure intuition.

We know that $\sin x$ alternates between $- 1$ and $1$, from negative infinity to infinity. We also know that $x$ increases from negative infinity to infinity. What we have, then, at large values of $x$ is a large number ($x$) multiplied by a number between $- 1$ and $1$ (due to $\sin x$).

This means the limit does not exist. We do not know if $x$ is being multiplied by $- 1$ or $1$ at $\infty$, because there is no way for us to determine that. The function will essentially alternate between infinity and negative infinity at large values of $x$. If, for example, $x$ is a very large number and $\sin x = 1$, then the limit is infinity (large positive number $x$ times $1$); but $\frac{3 \pi}{2}$ radians later, $\sin x = - 1$ and the limit is negative infinity (large positive number $x$ times $- 1$).