Simple two variable limit problem, find the limit, if exists, of $f(x,y)=x^2\sin(\frac{1}{xy}) $: 
Find $\displaystyle \lim_{(x,y)\to(0,0)} x^2\sin(\frac{1}{xy}) $ if exists, and find $\displaystyle\lim_{x\to 0}(\lim_{y\to 0} x^2\sin(\frac{1}{xy}) ), \displaystyle\lim_{y\to 0}(\lim_{x\to 0} x^2\sin(\frac{1}{xy}) )$ if they exist.

Hey everyone. I've tried using the squeeze theorem and found $0 \le |x^2\sin(\frac{1}{xy})| \le |x^2|\cdot 1 \xrightarrow{x\to0} 0 $ and so the "double" limit exists and equals zero.
Now, I know $\lim_{x\to 0}\sin(\frac{1}{ax})$ diverges, so both $lim_{y\to 0}(\lim_{x\to 0} x^2\sin(\frac{1}{xy}) )$  and $lim_{x\to 0}(\lim_{y\to 0} x^2\sin(\frac{1}{xy}) )$ do not exist(?) 
I don't think I understand multi-variable limits, I would love your help on this basic one so I can understand better. Thanks in advance :) 
 A: You only need simply to note that
$$0\le \left|x^2\sin(\frac{1}{xy})\right|\le x^2\to 0$$
then conclude by squeeze theorem.
A: $0 \le |x^2\sin(\frac{1}{xy})| \le |x^2| \xrightarrow{x\to0} 0$ means indeed that for any trajectory of $(x,y)$ going to $(0,0)$ we have that $x^2\sin(\frac{1}{xy})$ approaches $0$.

Note that just by taking any $y\neq 0$ fixed we cannot just conclude that it would not approach $0$, since eventually you also want $y$ to approach $0$. Indeed we can see that $y\neq 0$ fixed is a trajectory that does not intersect or approach $(0,0)$.
A: So the moral of this question is that, one might think that if the double limit exists, then it entails the existence of iterated limits, this question gives a counterexample, but what is true is that,

Given that $\lim_{(x,y)\rightarrow(a,b)}f(x,y)=L$ exists and that for each $x\in B_{\delta}(a)-\{a\}$, $\lim_{y\rightarrow b}f(x,y)=M_{x}$ exists, then $\lim_{x\rightarrow a}M_{x}=L$, in other words, $\lim_{x\rightarrow a}\lim_{y\rightarrow b}f(x,y)=\lim_{(x,y)\rightarrow(a,b)}f(x,y)$.

The assumption that for a deleted neigborhood of $a$ that the existence of $\lim_{y\rightarrow b}f(x,y)$ cannot be relaxed.
