What is the definition of a path along a multivariable function? I'm taking a class equivalent of Calculus III, and we saw how to prove continuity of a multivariable function. Recently we looked at the following example:
\begin{align}
 f(x,y) =
\begin{cases}
\frac{x^{2\alpha}}{x^2 + y^2},  & \text{if ($x$,$y$)} \neq \text{(0,0)} \\[1ex]
0 & \text{if ($x$,$y$) = (0,0)}
\end{cases}
\end{align}
Long story short, we ended up switching to polar coordinates and simplifying down to:
$$\lim_{r \to 0} r^{2\alpha - 2} \cdot \cos^{2\alpha}{\theta}$$
Besides the fact that we can now analyze different cases based on the value of $\alpha$, we came to the conclusion that since the point is approached at by infinitely different "paths" (depending on the value of $\theta$), the limit doesn't even exist in the first place. 
So, I get that this limit hinges on $\theta$ to some respects, but isn't it like a constant? I don't get why approaching the same point from different angles would result in no limit existing, does this have to do with the way curves are sometimes used to solve limits? 
(My visual knowledge on this kind of stuff is very limited, so I have a really hard time imagining how such a path would look like)
 A: Here is an example with your function. Fix $\alpha=1$ and imagine in rectangular coordinates, approaching the limit along the $x$-axis (so $y=0$). Then you get
$$
\frac{x^2}{x^2+y^2} \to \frac{x^2}{x^2+0} = 1.
$$
Now imagine approaching from the $y$-axis, so $x=0$. You get
$$
\frac{x^2}{x^2+y^2} \to \frac{0}{0+y^2} = 0.
$$
Since the values disagree, the limit does not exist.
This tells you that the notion of the limit (or convergence, if you like) in multiple dimensions is much more demanding than in 1-D.
A: The answer to your question hinges on what exactly $\alpha$ is supposed to be. If $\alpha$ is a parameter which defines a family of functions, there may be some values of $\alpha$ for which the function is continuous everywhere. It certainly is not continuous for all $\alpha$ in either $\mathbb{R}$ or $\mathbb{Z},$ since $\alpha = 0$ produces the function
$$
f = \begin{cases}
\frac{1}{x^2+y^2}\  & (x,y)\neq (0,0)\\
0 & (x,y) = (0,0)\\
\end{cases}
$$
which is not continuous at $(0,0)$. I suspect what you have proved is that there is generally no $\alpha$ for which this function is continuous everywhere, with $(0,0)$ being the point that spoils it. I'm sure someone else will correct me if I am wrong about this...
I think you may be confused between the transformation of coordinates and the transformation of a function. You transformed the coordinates, so you are coming in on the function from a different angle, yes, but it is the same function. The function is asymmetric for most $\alpha$, so it's unsurprising that you get different limits coming in from different angles. As you correctly argue, the failure of the limit to exist is a sign that the function is discontinuous at that point.
