Answer
0
Work Step by Step
We find:
\begin{array}{c}
x=r \cos \theta, \quad y=r \sin \theta \\
r^{2} =x^{2}+y^{2} \\
(x, y) \rightarrow(0,0) \quad \Rightarrow r \rightarrow 0+ \\
\left|\frac{x y}{x^{2}+2 y^{2}}\right|=\left|\frac{r^{2} \cos \theta \sin \theta}{\sqrt{r^{2}+2 r^{2} \sin ^{2} \theta}}\right| \leq \frac{r^{2}}{\sqrt{r^{2}}}=r \\
\lim _{r \rightarrow 0+} r=0 \quad \Rightarrow \quad \lim _{(x, y) \rightarrow(0,0)} \frac{x y}{x^{2}+2 y^{2}}=0
\end{array}
