Show that limit does not exist (two variables) I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.
I want to answer this question:

Show that the following limit does not exist
$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$

So, my working:
$$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\to 0}\frac{0}{y^2}=0$$
and
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{y\to 0}\frac{0}{x^2}=0$$
I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $y=x$, not sure if that's correct - or my working.
So, let $y=x$, then
$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2x^2+(x-y)^2}=\lim_{x\to 0}\frac{x^4}{x^4+(x-x)^2}=1$$
Therefore limit doesn't exist.
Is this somewhat correct? What's the best way to answer a question like this?
Also, when showing that this limit does not exist, do I need to find different values of limits for both ${x\to 0}$ AND ${y\to 0}$? Or is one enough, e.g. if I just find two different values for two limits for ${x\to 0}$ without using ${y\to 0}$ in my calculation at all, is that fine?
Thanks!
 A: What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same. 
So when you calculate $$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.
Similarly,
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.
And the last limit you calculated is along line $y=x$.
So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.
A: Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.
A: What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.
A: What you did is completely correct, but perhaps you can ease it a little bit as follows:
For $\;x=0\;,\;\;y\to0\;$ the limit clearly is $\;0\;$, whereas for $\;x=y\;$ and $\;x\to0\;$ we get
$$\frac{x^4}{x^4+(x-x)^2}=1\xrightarrow[x=y\to0]{}1$$
Thus the limit depends on the path chosen $\;\implies\;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.
A: You did right.
A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
$$
\lim_{x\to0}\frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
\lim_{x\to0}\frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
\lim_{x\to0}\frac{m^2}{m^2+(1-m)^2x^{-2}}=
\begin{cases}
0 & m\ne 1 \\[4px]
1 & m=1
\end{cases}
$$
Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.
