Does $\lim_{x\to 0}\sin(x)\sin(1/x)$ exist?

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?

• It exists and is $0$ because $$\lvert\sin(1/x)\rvert\le1$$ and $$\sin(x)\xrightarrow{x\to0}0$$ – Maximilian Janisch Dec 26 '19 at 18:33
• 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). – Ted Shifrin Dec 26 '19 at 19:07

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 $$\lim_{x \to 0} \sin (x) \sin (1/x) = 0$$.

• 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 $0. \qquad$ – Michael Hardy Dec 27 '19 at 7:34

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.

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 $$x\to 0.$$

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.