Showing $\lim_{(x,y) \to (0,0)} xy \log(x^2+y^2) = 0$ First I let $x=r\cos \theta, y = r\sin \theta$ and so limit
$$\lim_{r\to 0} r^2\sin2\theta \log(r)$$
Now, in region $0<x<1$, $\log(x) < 1/x$ 
$$|r^2\sin 2\theta \log (r) - 0| < |r\sin 2\theta| \le |r| < \delta < \epsilon$$
So limit exist if $\delta < \epsilon$ and limit is 0.

Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 \log r$ as $\log(r) / (r^{-2})$ which again gave 0.
 A: 
I thought it might be instructive to present an approach that is direct and follows from a pair of elementary inequalities only.   To that end we proceed.

Using $\color{blue}{|\log(x)|<\frac1{\sqrt{x}}}$ for $x\le 1$ along with $\color{red}{|xy|\le \frac12(x^2+y^2)}$, we have for any given $\varepsilon>0$
$$\begin{align}
\left|\color{red}{xy}\color{blue}{\log(x^2+y^2)}\right|&\le \color{red}{\frac12 (x^2+y^2)}\color{blue}{\frac1{\sqrt{x^2+y^2}}}\\\\
&=\frac12 \sqrt{x^2+y^2}\\\\
&<\varepsilon
\end{align}$$
whenever $\sqrt{x^2+y^2}<\delta=2\varepsilon$.
And we are done!
A: We have
$$xy \log(x^2+y^2) =(x^2+y^2)\log(x^2+y^2)\cdot \frac{xy}{x^2+y^2}\to 0$$
indeed since $t=x^2+y^2\to 0$
$$(x^2+y^2)\log(x^2+y^2)=t\log t\to 0$$
and since $x^2+y^2\ge 2xy$
$$0\le \left|\frac{xy}{x^2+y^2}\right| \le \frac12$$
A: More easily, $lim_{r\rightarrow 0}rlog(r)=0$, we deduce that $lim_{r\rightarrow 0}r^2log(r)=0$ since $|sin(\theta)|\leq 1$, the result follows.
Limit of $x \log x$ as $x$ tends to $0^+$
A: First, notice that from the (in)equalities $0\leq (x-y)^2=x^2+y^2-2xy$ and $\log(r)\leq r-1$ there holds that
$$
0 \leq \lim_{(x,y)\rightarrow (0,0)} |xy\log(x^2+y^2)| \leq \lim_{(x,y)\rightarrow (0,0)}\frac{1}{2}(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
