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+ \\
y \ln \left(x^{2}+y^{2}\right)=r \sin \theta \ln r^{2}=2 r(\ln r) \sin \theta \\
\lim _{(x, y) \rightarrow(0,0)} y \ln \left(x^{2}+y^{2}\right)=\lim _{r \rightarrow 0+} 2 r(\ln r) \sin \theta=2 \sin \theta\left[\lim _{r \rightarrow 0+} r(\ln r)\right]=\ldots \\
\lim _{r \rightarrow 0+} r(\ln r)=\lim _{r \rightarrow 0+} \frac{\ln r}{\frac{1}{r}}=\left(\mathrm{L}^{\prime} \mathrm{H}, \frac{\infty}{\infty}\right)=\lim _{r \rightarrow 0+} \frac{\frac{1}{r}}{-\frac{1}{r^{2}}}=\lim _{r \rightarrow 0+}-r=0
\end{array}
