Q:

Limit definition (2 variables)

I have the following function:
$$ f(x,y) = \begin{cases} 
      \dfrac{xy^3}{{x^2+y^2}} & (x,y)\neq(0,0) \\
      0 & (x,y)=(0,0) \\ 
\end{cases} $$
How can I demonstrate the existence of such a limit using epsilon and delta?

A:

Let $\varepsilon > 0$.
Choose $\delta = \sqrt{\varepsilon}$.
For $\|(x,y)\|_2 < \delta$ we have:
$$\left|\frac{xy^3}{x^2 + y^2}\right| = \frac{|x||y|^3}{\|(x,y)\|_2^2} = \underbrace{\frac{|x|}{\|(x,y)\|_2}}_{\le 1} \cdot \underbrace{\frac{|y|}{\|(x,y)\|_2}}_{\le 1} \cdot\underbrace{|y|^2}_{<\delta^2} < \delta^2 = \varepsilon$$
since $\max\{|x|,|y|\} \le \|(x,y)\|_2$.
Thus $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy^3}{x^2 + y^2} = 0$ so $f$ is continuous at $(0,0)$.

A:

Presumably, the limit you want to check is $$\lim_{(x,y)\to(0,0)} f(x) = \lim_{(x,y)\to(0,0)} \frac{xy^3}{x^2+y^2}$$
There are some useful inequalities that help with questions like this. One comes from noting the following. Since $(x+y)^2\ge 0$, and $(x-y)^2\ge 0$, we can expand and rearrange these to conclude that $|xy|\le x^2+y^2$, for all $x$ and $y$. (In fact, you get that $2|xy|\le x^2+y^2$, but we can use the weaker version.)
Now, you can rewrite the expression in question as $\frac{xy}{x^2+y^2}\cdot y^2$. By the inequality derived above, the first fraction is always $\le 1$ in absolute value. Then, we know that $y^2\to 0$ as $(x,y)\to (0,0)$. To set up your $\epsilon-\delta$ argument, you just need to know that $y^2<\epsilon$, which will happen if $y<\sqrt{\epsilon}$.
Setting $\delta=\sqrt{\epsilon}$, we can now argue that $|f(x,y)|<\epsilon$ under various circumstances, such as $\max\{|x|,|y|\}<\delta$, or $\sqrt{x^2+y^2}<\delta$, or $|x|+|y|<\delta$, any of which puts us in a small neighborhood of the origin.

