Answer
0
Work Step by Step
\[
\lim _{(x, y) \rightarrow(0,0)} \sqrt{x^{2}+y^{2}} \ln \left(x^{2}+y^{2}\right)
\]
Convert to polar coordinates
\[
x^{2}+y^{2}=r^{2} \text { and } \theta=\tan ^{-1}\left(\frac{y}{x}\right)
\]
$x^{2}+y^{2}=r^{2} \Rightarrow r=\sqrt{x^{2}+y^{2}}$
If $r=\sqrt{x^{2}+y^{2}}$ and $(x, y) \rightarrow(0,0)$ then $r \rightarrow 0$
Therefore, $\underbrace{\lim _{(x, y) \rightarrow(0,0)} \sqrt{x^{2}+y^{2}} \ln \left(x^{2}+y^{2}\right)}_{\downarrow}$
\[
\lim _{r \rightarrow 0} r \ln \left(r^{2}\right)
\]
The limit can be written as
\[
\begin{array}{l}
\lim _{r \rightarrow 0} \frac{\ln r^{2}}{\frac{1}{r}} \\
\lim _{r \rightarrow 0} \frac{2 \ln r}{\frac{1}{r}}
\end{array}
\]
Now apply the L'Hopital's rule
$\lim _{r \rightarrow 0} \frac{\ln r^{2}}{\frac{1}{r}}=\lim _{r \rightarrow 0} \frac{2 \frac{d}{d t}[\ln r]}{\frac{d}{d r}\left[\frac{1}{r}\right]}$
$\lim _{r \rightarrow 0} r \ln \left(r^{2}\right)=\lim _{r \rightarrow 0} \frac{2\left(\frac{1}{r}\right)}{-\frac{1}{r^{2}}}$
$\lim _{r \rightarrow 0} r \ln \left(r^{2}\right)=-\lim _{r \rightarrow 0} \frac{2 r^{2}}{r}=-\lim _{r \rightarrow 0} 2 r$
Evaluate the limit
$-\lim _{r \rightarrow 0} 2 r=-2(0)$
$-\lim _{r \rightarrow 0} 2 r=0$
Therefore $\lim _{(x, y) \rightarrow(0,0)} \sqrt{x^{2}+y^{2}} \ln \left(x^{2}+y^{2}\right)=0$
