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I want to calculate

  • $\lim_{x\to\ 0} \sin(x)\sin(1/x)$

But I have to calculate both sides since $1/x$ is not defined for $0$.

  • $\lim_{x\to\ 0+} \sin(x)\sin(1/x)$

  • $\lim_{x\to\ 0-} \sin(x)\sin(1/x)$

And I wonder whether it exists, because that $\sin(1/x)$ does not exist and $\sin(x)$ is zero, so zero * does not exist means that this limit on each side does not exists?

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    $\begingroup$ It exists and is $0$ because $$\lvert\sin(1/x)\rvert\le1$$ and $$\sin(x)\xrightarrow{x\to0}0$$ $\endgroup$ – Maximilian Janisch Dec 26 '19 at 18:33
  • $\begingroup$ Remember the most basic thing about limits: Whether things are or are not defined at the limiting point (here, $x=0$) is totally irrelevant. Only in the uninteresting cases do you literally plug in the limiting value of $x$ (and this is when you have a continuous function). $\endgroup$ – Ted Shifrin Dec 26 '19 at 19:07
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We can use the squeeze theorem here. First, can you prove that for all $x$ other than $0$,

$$-|x| \le \sin (x) \sin (1/x) \le |x|?$$

After you've shown that, since $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$, the squeeze theorem tells us that $\lim_{x \to 0} \sin (x) \sin (1/x) = 0$.

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  • $\begingroup$ I'd phrase it a bit more simply: $\sin(1/x)$ remains between $+1$ and $-1,$ so $(\sin x) (\sin(1/x))$ remains between $+\sin x$ and $-\sin x,$ and those both approach $0. \qquad$ $\endgroup$ – Michael Hardy Dec 27 '19 at 7:34
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The limit does exist. Focus on the $\sin\frac{1}{x}$ part.

It has an oscillating discontinuity at $x=0$. The function oscillates between $[-1,1]$ with greater frequency the closer $x$ gets to zero. (Is it clear why?)

However, $\sin x\to 0$ as $x\to 0$. Therefore, a finite quantity multiplied by zero will yield zero.

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The limit exists and is $0,$ same as the limit of the multiplier $\sin x.$

Forget for the moment about $\sin x$ and use the multiplier $x$ instead, then if you can see that $$x\sin\left(\frac 1x\right)$$ has a limiting value at $x=0$ which is $0,$ then you should be able to see that this same line of thought essentially unchanged applies to the function we get by replacing $x$ with $\sin x,$ namely $$\sin x\sin\left(\frac 1x\right),$$ which has the same limit $0$ as $x\to 0.$

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There is a really useful theorem that you can use in general:

Let $l(x)$ be a function that is limited in a neighborhood of $x_0$ and let $i(x)$ be infinitesimal as $x \to x_0$, then: $$\lim_{x\to x_0} l(x)i(x)=0$$

The proof is pretty simple and uses squeeze theorem(or 2 carabinieri theorem as we call it in Italy ☺️) and some elementary absolute value results.

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