# Finding the limit for functions with two variables

I know that when we have a limit of a function with $2$ variables, the limit must be the same, regardless of the path we take. So this is useful for proving that a limit doesn not exist. But when you've tried this method for different "paths" (e.g., $(x,0),(0,y),(x,x),(y,y)$, etc...) and you think that the limit does exist, how you do show it?

For example here is a question from my textbook:

Evaluate the limit: $$\lim_{\large{(x,y) \to (0,0)}} \dfrac{xy \sin(xy)}{x^2+y^2}$$

The answer is supposed to be $0$ but I don't see how you can prove that it is $0$ for any direction you approach $(0,0)$ from. Can someone please help me? Thanks.

• Do you know how to convert these limits to polar coordinates? That can be at times a fast and easy way to prove a limit exists. mrf's hint is also very helpful in applying the squeeze theorem. – Jared May 16 '13 at 5:40
• @Jared I think I should know that. I remember my teacher saying something about polar coordinates but I've never actually used it. Could you give some pointers? Or a link to some site would be helpful. Thanks. – Jey May 16 '13 at 5:43
• Thanks for the latex edit. – Jey May 16 '13 at 5:54

If you change to polar coordinates, you get $$\lim_{r \rightarrow 0} \sin(\theta) \cos(\theta) \sin(r^2 \sin(\theta) \cos(\theta)).$$

It might look a bit confusing, but the essential thing is that $\sin(x)$ goes to zero as $x$ goes to zero, and that $\sin(\theta)$ and $\cos(\theta)$ are at most 1.

• I understand how you changed the expression using the substitutions $x=r\cos{\theta}$ and $y=r\sin{\theta}$. But how do change $(x,y) \rightarrow (0,0)$ to $r \rightarrow 0$? Thanks. – Jey May 16 '13 at 5:49
• That $(x,y) \rightarrow 0$ means that the distance from the point $(x,y)$ to $(0,0)$ should go to zero, which means that $\sqrt{x^2 + y^2} = r$ should go to zero. The $\theta$ varies freely, but this is ok for us since the expression goes to zero. Note that for instance $\lim_{r \rightarrow 0} \sin(\theta)$ does not exist. – N.U. May 16 '13 at 6:03

Hint: $\left|\dfrac{xy}{x^2+y^2}\right| \le \dfrac12$.

• How do you get 1/2? – Jey May 16 '13 at 5:47
• @Jey: $x^2+y^2-2xy \ge 0 \implies x^2+y^2 \ge 2xy \implies \dfrac{1}{x^2+y^2} \le \dfrac{1}{2xy}$ – Inceptio May 16 '13 at 5:49
• AM/GM inequality (or $\sin \theta \cos \theta = \frac 12 \sin 2\theta$) – mrf May 16 '13 at 5:49
• Nice. I didn't notice that. Thanks – Jey May 16 '13 at 5:51