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.
 A: $$0<\sqrt{x^2+y^2}<\delta\iff 0<x^2+y^2<\delta^2$$
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.
A: We have $$\frac{x^2y^2}{x^2+y^2}=\frac{1}{\dfrac{1}{x^2}+\dfrac{1}{y^2}}.$$ Note that each term in the denominator on the RHS is positive and independently goes to $\infty$ as $x$ and $y$ approach $0$. So the required limit is $0$.
A: Given $\varepsilon>0$, take $\delta=\sqrt{\varepsilon}$. Then,
\begin{align*}
0<\sqrt{x^2+y^2}<\delta\quad&\Rightarrow \quad x^2<\delta^2\text{ and }y^2<\delta^2\\
&\Rightarrow \quad\frac{1}{x^2}+\frac{1}{y^2}>\frac{2}{\delta^2}\\
&\Rightarrow \quad \left|\frac{x^2y^2}{x^2+y^2}-0\right|=\frac{1}{\frac{1}{y^2}+\frac{1}{x^2}}<\frac{1}{\frac{2}{\delta^2}}=\frac{\varepsilon}{2}.
\end{align*}
A: We have:
$$0\leq\frac{x^2y^2}{x^2+y^2}\leq \frac{x^2y^2}{y^2}=x^2$$
Since $f(x,y)$ is between $0$ and $x^2\rightarrow 0$, the function is arbitrarily close to zero, when the distance between $(x,y)$ and $(0,0)$ is sufficiently small. It follows that the limit of $f$ is $0$, just to limit definition.
