Calculate double limit of $x^2\sin\frac{1}{xy}$

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.

• But using polar coordinates will give the same limit if it exists. I know that if the last one exists, then the first two exist and equal, hence I wanted to found first whether it exists. How should I tackle the question? – Galc127 Mar 24 '15 at 6:36
• @kennytm Both statements you said are wrong. The last limit can be evaluated by going to polar coordinates. It is only a change of variable. Also, the existence of the last limit doesn't ensure the existence of the first two limits. In this example, what happens is that $f(x)=\lim_{y\to0}x^2\sin(1/xy)$ is not defined in a neighborhood of $x=0$. Therefore, $\lim_{x\to0}f(x)$ doesn't make sense. – Nathanson Mar 24 '15 at 7:15
• @Nathanson, you are wrong. It's a theorem that if the last limit exists, than the first two exist and equal. On the other side, if the two first not exist or not equal, the last one not exist. Your last claim is what I needed. Thanks. – Galc127 Mar 24 '15 at 7:19
• @Galc127 No, you need to review the theorem. The theorem requires the existence of the last limit and the existence of $\lim_{y\to}f(x,y)$ in a neighborhood of $x=0$. – Nathanson Mar 24 '15 at 7:21
• @Nathanson, understood. Thanks! – Galc127 Mar 24 '15 at 7:26

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 $x=0$.

• I don't understand why $\displaystyle 0\le x^2\sin\frac{1}{xy}$. – Galc127 Mar 24 '15 at 7:27
• @Galc127 I forgot to type the absolute values. – Nathanson Mar 24 '15 at 7:29