Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta$, and I cannot see a way to obtain $\sqrt{x^2+y^2}$ in the above inequality to complete the proof. Am I mistaken in my process? Thank you.

$$0<\sqrt{x^2+y^2}<\delta\iff 0<x^2+y^2<\delta^2$$
But you have already shown that $\left\lvert\frac{x^2y^2}{x^2+y^2}-0\right\rvert<x^2+y^2$ and so $$\implies\left\lvert\frac{x^2y^2}{x^2+y^2}-0\right\rvert<\delta^2$$
So choosing $\delta=\sqrt{\epsilon}$ suffices.
