Find limits

Quote:

Originally Posted by Xingyuan

$\lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)$

How to find the above limits?

using polar coordinates we have $y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.$ since $(x,y) \to (0,0),$ we have $r \to 0+$ and thus, as a result, $\ln r < 0.$ hence $2r^2 \ln r < 2r^2 \sin^2 \theta \ln r < 0.$

finally, since $\lim_{r\to0+} 2r^2 \ln r = 0,$ the squeeze theorem shows that the required limit is 0.