Proving that the function $f(x,y)=\frac{x^2y}{x^2 + y^2}$ with $f(0,0)=0$ is continuous at $(0,0)$. How would you prove or disprove that the function given by
$$f(x,y) = \begin{cases} \dfrac{x^2y}{x^2 + y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$
is continuous at $(0,0$)?
 A: 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.
A: 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.
A: HINT
I would use polar coordinates, and consider $r \to 0$
