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.
Best Answer
To answer your question expressly, in the second line, you're multiplying by $2$ explicitly, and then dividing by $2$ implicitly by taking the natural logarithm of the square root of the original quantity. The claim, then, is that $\dfrac{\vert x+y \vert}{\sqrt{x^2+y^2}} \leq \sqrt 2$. As Kavi Rama Murthy notes, that's true via Cauchy-Schwarz.
Once you know that $f(x, y)$ is bounded above by a constant times $r \ln r$, you're done because it's easy to see (via L'Hopital, for instance) that the latter expression goes to $0$ as $r \to 0^+$.