Limit of $\sin\left(\frac{\pi}{x} \right)$ as $x$ approaches $0$ does not exist, with squeeze theorem I understand that $\frac{\pi}{0}$ would be infinity, hence $\sin\left(\frac{\pi}{x} \right)$ does not have a limit as $x$ approach $0$.
But if I would to use the squeeze theorem:
$$-1 \leq \sin\left(\frac{\pi}{x} \right) \leq 1$$
shouldn't it be zero? Or does the squeeze theorem not work here?
 A: Just because it is bounded doesn't mean it has a limit. For example, I can construct a sequence: $$-1,1,-1,1,-1,1,\dots$$
and for this sequence, it is true that $-2\leq a_n \leq 2$. That only proves that the sequence is bounded, not that it converges.
In order for the squeeze theorem to work, the left and right sides of the inequality need to converge. For example, if $f(x)=\frac{\sin x}{x}$ and $x\to\infty$, then you could say $$-\frac 1x<f(x)<\frac1x,$$
and because $$\lim_{x\to\infty}\frac1x=\lim_{x\to\infty}-\frac1x=0$$
the limit of $f$ is also $0$.
You don't have that. You only know that $-1\leq f(x)\leq 1$ which is not enough to use the squeeze theorem.

That said, another huge warning:
Your argument, saying

$\pi/0$ would be infinity, hence $\sin(\pi/x)$ does not have a limit as $x\to0$

is completely inadequate. There are several functions $f$ for which $\lim_{x\to 0} f(\frac{x}{\pi})$ exists, even though, as you say, "$\frac\pi0$ would be infinity". For example, take $f(y) = e^{-|y|}$. Then the limit $$\lim_{x\to 0} f(\frac\pi x) = \lim_{x\to 0}e^{-\frac{\pi}{|x|}}$$
exists and is equal to $0$.
