calculus - Calculate double limit of $x^2sinfrac{1}{xy}$

I'm trying to compute the follow limits:

$\displaystyle\lim_{x\to 0}\lim_{y\to 0} x^2\sin\frac{1}{xy} \ , \ \lim_{y\to 0}\lim_{x\to 0} x^2\sin\frac{1}{xy} \ , \ \lim_{(x,y)\to (0,0)}x^2\sin\frac{1}{xy}$

First, I tried to compute the last one.
$$\lim_{(x,y)\to (0,0)}x^2\sin\frac{1}{xy}=\lim_{r\to 0}r^2\cos^2 \theta\sin\frac{2}{r^2\sin(2\theta)}$$
I tried to use squeeze theorem and got $0$. According to W|A the limit doesn't exist and I don't understand why.

The middle one is pretty easy - using squeeze theorem, the inside limit is $0$, thus the whole limit is $0$.

I don't how to start evaluating the first one.

Please help, thank you.

Answer

The last limit exists:

$$0\leq |x|^2\cdot\left|\sin\left(\frac{1}{xy}\right)\right|\leq |x|^2\to0$$

Therefore $$\lim_{(x,y)\to(0,0)}x^2\sin\left(\frac{1}{xy}\right)=0.$$

The second limit exists:

It is the same argument. We compute by squeeze that $g(y)=\lim_{x\to0}x^2\sin\left(\frac{1}{xy}\right)=0$, for $y=\neq0$, and therefore $$\lim_{y\to0}\lim_{x\to0}x^2\sin\left(\frac{1}{xy}\right)=\lim_{y\to0}g(y)=\lim_{y\to0}0=0$$

The first limit doesn't exist make sense:

The problem is that

$$f(x)=\lim_{y\to0}x^2\sin\left(\frac{1}{xy}\right)$$

doesn't exist for $x\neq0$. Therefore

$$\lim_{x\to0}\lim_{y\to0}x^2\sin\left(\frac{1}{xy}\right)=\lim_{x\to0}f(x)$$

doesn't make sense. To define limit we need the function to be defined in a neighborhood of $x=0$, or at least in a set that accumulates at $x=0$.