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