Use Epsilon Definition of Limit to Prove Limit Exists We have the function
$$f(x,y)=
\begin{cases}
0  & \textrm{if } x=y=0 \\[2ex]
(x+y)\ln(x^2+y^2) & \textrm{otherwise}
\end{cases}$$
and need to prove  $\lim\limits_{(x,y) \to (0,0)} f(x,y) = 0$.
We need to show that for any $\epsilon>0$ we can find a $\delta > 0$ such that for any $(x.y)$ with $0< \sqrt{x^2+y^2}<\delta$, we have $|f(x,y)| < \epsilon$
Some of the steps are:
$$\begin{align}|f(x,y)| &= \left|(x+y)\ln(x^2+y^2)\right| \\&= \left|\frac{2(x+y)}{\sqrt{x^2+y^2}}\sqrt{x^2+y^2}\ln(\sqrt{x^2+y^2})\right| \\&\le 2\sqrt{2}\left|\sqrt{x^2+y^2}\ln(\sqrt{x^2+y^2})\right|
\end{align}$$
but I do not understand how you get including expression including $\sqrt{x^2+y^2}$ after plugging in the value to the absolute value and how that relates to the inequality. There has to be a step I am missing.

You can actually skip the second equality and note that $ |x+y| \leq \sqrt {1^{2}+1^{2}} \sqrt {x^{2}+y^{2}}$ by Cauchy-Schwarz inequality.
