Find $\lim _{x,y \to 0} x^2 y^2\ln (x^2+y^2)$

Question

How to find $\lim _{x,y \to 0}x^2y^2 \ln (x^2+y^2)$? Could $x^2y^2 \ln (x^2+y^2)=\frac{\ln(x^2+y^2)}{x^2+y^2}x^2y^2(x^2+y^2)$ be the way?

Answer 1

We have that

$$x^2 y^2\ln (x^2+y^2)=\frac{x^2 y^2}{x^2+y^2}\cdot(x^2+y^2)\cdot\ln (x^2+y^2) \to 0$$

indeed by standard limits for $u\to 0 \implies u\log u \to 0$ we have that

$$(x^2+y^2)\cdot\ln (x^2+y^2) \to 0$$

and by polar coordinates

$$\frac{x^2 y^2}{x^2+y^2}=r^2\sin^2\theta\cos^2\theta\to 0$$