$\displaystyle \lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)$
How to find the above limits?
using polar coordinates we have $\displaystyle y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.$ since $\displaystyle (x,y) \to (0,0),$ we have $\displaystyle r \to 0+$ and thus, as a result, $\displaystyle \ln r < 0.$ hence $\displaystyle 2r^2 \ln r < 2r^2 \sin^2 \theta \ln r < 0.$
finally, since $\displaystyle \lim_{r\to0+} 2r^2 \ln r = 0,$ the squeeze theorem shows that the required limit is 0.