Question

find the limit (if it exists). (if an answer does not exist, enter dne.)
\\(\lim_{(x,y)\to(0,0)}\ln(x^{2}+y^{2})\\)

Explanation:

Step1: Convert to polar coordinates

Let $x = r\cos\theta$ and $y = r\sin\theta$. Then $x^{2}+y^{2}=r^{2}$, and as $(x,y)\to(0,0)$, $r\to0$. The limit becomes $\lim_{r\to0}\ln(r^{2})$.

Step2: Simplify the expression

$\lim_{r\to0}\ln(r^{2})=\lim_{r\to0}2\ln(r)$. As $r\to0^{+}$, $\ln(r)\to-\infty$, so $2\ln(r)\to-\infty$.

Answer:

$-\infty$