# Proving that the function $\frac{x^2y}{x^2 + y^2}$ is continuous at $(0,0)$.

How would you prove or disprove that the function given by $$f(x,y) = \begin{cases} \frac{x^2y}{x^2 + y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$ is continuous at $(0,0$)?

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– user20266 Jun 8 '12 at 14:11

Observe that $$\left| \frac{x^2y}{x^2+y^2} \right| \leq \frac{x^2 |y|}{x^2} = |y|$$ provided $x \neq 0$ and $y \neq 0$. Then you conclude.

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HINT

I would use polar coordinates, and consider $r \to 0$

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Elegant complete hint. – S. Snape Jun 8 '12 at 14:08

A function is continuous at a point if the limit exists there and the function takes that value at that point. A limit exists iff it has the same value no matter how the point is approached. In functions of one variable this is easy to test, since you only have 2 options. In more variables it's harder, because it must work for any path to the point. Not only must you then consider every angle, but you must consider every path. You could get the same limit approaching on any straight line, but not along some more winding curve.

The important thing to realize is that the $(x,y)$ form of the function hides a useful fact that is made clear in polar form $(r,\theta)$. All paths to the origin have $r\rightarrow 0$, so this is what they have in common. That means that what makes them different is entirely in how $\theta$ changes as you approach the origin. So a simple trick is to convert to polar form and see if the limit depends on $\theta$ in any way. If it does, then the limit doesn't exist. If it doesn't, then the limit pops right out. If that limit is the same as the value of the function at that point ($0$ here), then the function is continuous there.

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"...see if the resulting expression depends on $\theta$ in any way. If it does, then the limit doesn't exist..." I find this very unclear. Of course the expression you are taking a limit of can depend on $\theta$ even when the limit exists (e.g. it is the case here), and I'm sure this isn't what you mean. – Jonas Meyer Jun 8 '12 at 15:10
@JonasMeyer thank you. Yes, I meant when the limit itself depends on $\theta$. – Robert Mastragostino Jun 8 '12 at 16:01
Thank you, but I still don't think it is entirely clear what you mean. Referring to depedence on $\theta$ of the limit makes it sound like you are first fixing $\theta$ in order to let $r\to 0$, but this would only give the radial limits, which might all be equal wittout the limit existing, as you more or less indicate in your first paragraph. – Jonas Meyer Jun 8 '12 at 16:11
What I mean to say is to take the limit as $r\rightarrow 0$, and see if the value of the resulting expression after that depends on $\theta$. If it doesn't, then it's clearly the value of the limit. Is this a good enough wording to include, or do I still sound ambiguous? – Robert Mastragostino Jun 8 '12 at 16:16
If you actually take the limit as $r\to 0$, then it cannot possibly have a value that depends on $\theta$. It will either exist, in which case it will be a number, or it will not exist. By the way, I think that your answer is generally helpful and upvoted it before leaving my first comment. – Jonas Meyer Jun 8 '12 at 16:26