# How to proof $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^4}$ using the $\epsilon-\delta$ definition?

I am stuck on proving that the limit $L = \lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^4}$ does or does not exist (using the definition $\forall\epsilon\in\mathbb R_0^+:\exists\delta\in\mathbb R_0^+:\forall(x, y)\in\mathbb R^2-(0,0):||(x,y)||<\delta\Rightarrow|\frac{x^2y^2}{x^2+y^4}-L|<\epsilon$. I've found that if the limit exists, then $L$ must be equal to $0$ (by approaching the limit on several paths like $y=x$). I haven't found a path were the limit is not equal to 0, so I assume the limit does exist, but I can't find a way to proof the limit using the $\epsilon-\delta$ definition.

Just notice that $0 \leq \frac{x^2y^2}{x^2 + y^4} \leq \frac{x^2y^2}{x^2} = y^2$ since $y^4 \geq 0$. Now you just need to choose a $\delta > 0$ small enough (using the euclidic norm might help you the most) so that $\vert y^2 \vert < \epsilon$ and you are done :)
hint: $0 \le \dfrac{x^2y^2}{x^2+y^4} \le \dfrac{x^2y^2}{2|x|y^2} = \dfrac{|x|}{2}$. Use this double inequality together with the definition you can find the "$\delta$" in terms of "$\epsilon$" and completes the proof in elegant manner.
• Where did you get $x^2+y^4\ge2\lvert x\rvert y^2$? – Akiva Weinberger Dec 30 '16 at 13:53
• Because $(|x|-y^2)^2 ≥ 0$, so we get $x^2-2|x|y^2+y^4 ≥ 0$, then we find $x^2+y^4 ≥ 2|x|y^2$. – mitchbus Dec 30 '16 at 14:04
$$0\leq{x^2y^2\over x^2+y^4}={|x|\over2}\>{2|x|y^2\over x^2+y^4}\leq{|x|\over2}\ .$$