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!
Best Answer
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.