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Evaluate: $$\lim_{x \rightarrow \infty}x\sin \frac{1}{x}$$

$$\lim_{x \rightarrow \infty}x \times\lim_{x \rightarrow \infty} \sin \frac{1}{x}=\infty \times0=\text{Undefined}$$ Is this the correct way to convey that the limit does not exist? Or is there a mathematical way to show that $$\lim_{x \rightarrow \infty} \sin \frac{1}{x} = 0$$ Other than just knowing that 1 divide by an infinitely large number approaches $0$.

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    $0\times\infty$, as it arises in a limit, is not so much "undefined" as it is "indeterminate." The limit may exist, or it may not. More work (L'Hospital's rule, e.g.) is required to see what's going on. – 2017-02-17
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    [This is how it looks](https://www.wolframalpha.com/input/?i=plot+x*sin(1%2Fx)+x+from+0+to+2) – 2017-02-17

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You can only use the fact that $$\lim_{x \to \infty}f(x)g(x)=\lim_{x \to \infty} f(x) \lim_{x \to \infty}g(x)$$ When it is given both limits exist. So your method of saying this is undefined is incorrect.

So the way to evaluate this limit is by setting $\frac{1}{x}=t$, which yields that $$\lim_{x \to \infty} x\sin \frac{1}{x} =\lim_{t \to 0^{+}} \frac{\sin t}{t}=1$$ As is discussed here.

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    What about the x before sin? is that not equal to infinity? – 2017-02-17
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    @NickPavini the fact that it is infinity doesn't matter. That is like saying $$\lim_{x \to 0} \frac{x}{x}$$ does not exist because the $\frac{1}{x}$ term goes to $\infty$. – 2017-02-17
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    @NickPavini You get indeterminate froms... – 2017-02-17
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    Ok I really like what you did, just having a hard time wrapping my head around it. – 2017-02-17
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    @NickPavini Thank you for your complements. :) You can ask further questions here as well. – 2017-02-17
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    This might be a dumb question but what happens to the $x$. I see how setting $$\frac{1}{x}=t$$ helps but a little lost on where $x$ goes. If that makes any sense. – 2017-02-17
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    @NickPavini $$ \frac{1}{x}=t \implies x=\frac{1}{t} $$ $$x \sin \frac{1}{x}=\frac{1}{t} \sin t$$ – 2017-02-17
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    That could not have been more of a perfect explanation! Thank you. – 2017-02-17
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    Last question however, why is it as $$x \rightarrow 0^+$$ instead of just as $$x \rightarrow 0$$? – 2017-02-17
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    @NickPavini $$\lim_{x \to \infty} \frac{1}{x}$$ Goes to $0$ from numbers bigger than $0$. (a bit of an handwavy explanation, but I think you'll understand). – 2017-02-17
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    definitely, Thanks again. – 2017-02-17
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    Some may argue that transforming the "question" $\lim_{x \to \infty}x\sin \frac{1}{x}$ into $\lim_{t \to 0} \frac{\sin t}{t}$ does not really make the problem any easier. However, if that latter limit is "well-known" or already treated earlier, it makes sense, of course. (We may appeal to the differential quotient of sine, but such a proof could be a circular argument, depending on the order we choose to prove things in). – 2017-02-18
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    @JeppeStigNielsen Yes, the latter limit is well known to the OP. I think from the OP's reaction he has already seen it. But perhaps I would include a link. – 2017-02-18
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This doesn't actually demonstrate that the limit doesn't exist; it demonstrates that the simplest possible 'rule' for establishing the limit won't work.

The key thing here is to realize that this is the same as $$ \lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\lim_{w\to0^+}\frac{\sin w}{w}=\lim_{w\to0^+}\frac{\sin w -\sin 0}{w-0}, $$ which is a special form you may recognize.

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For $x$ big enough, $\tfrac1x$ is close to zero and $\sin \frac1x=\frac1x+O(\frac1{x^3})$. Then $$ x\sin\frac1x=1+O(\frac1{x^2})\xrightarrow[x\to\infty]{}0 $$