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The question asks to evaluate the limit and discuss the continuity of the function. I think I made steps in the right direction, but I'm not sure how to go from there

$$\lim_{x, y\to(0, 0)} \frac{x^2y}{x^4+y^2}?$$

I divided the numerator and the denominator by $x^2y$ and got $$\lim_{x, y \to(0, 0)} \frac{1}{\frac{x^2}{y}+\frac{y}{x^2}}?$$

I was thinking of setting $a=\dfrac{x^2}{y}$ and $\dfrac 1a=\dfrac{y}{x^2}$. However, I'm not sure what $\lim_{x, y\to(0, 0)} a$ is. I suspect it's $0$, but I'm not sure. I'm also not sure if I even have to use the epsilon delta definition of a limit to prove that it exists before solving it.

Martin Sleziak
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Ovi
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1 Answers1

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Hint: try $y = x$ and $x = \sqrt{|y|}$ and look at the limiting value in each case.

K. Miller
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  • I got $0$ for one case and $1/2$ for the other so it worked :) but is there a more general way to do it? – Ovi Oct 12 '15 at 13:44
  • If you want to prove the limit doesn't exist, then this approach is usually sufficient. If you suspect the limit does exist, then you need an epsilon-delta proof. – K. Miller Oct 12 '15 at 14:56