What is the limit of $x sin x$ $xsinx$ as x approaches infinity?

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What is the limit of $x sin x$ $xsinx$ as x approaches infinity?

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Answer 1 · 2016-07-18T13:23:35

The limit does not exist. See below.

Explanation:

We can determine the result by pure intuition.

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

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

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What is the limit of $x sin x$ $xsinx$ as x approaches infinity?

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