Limit of $y \ln (x^2+y^2)$

Question

I want to calculate limit of $\lim_{(x,y) \rightarrow(0,0)}y \ln (x^2+y^2)$. How to do that? From iterated limits i know that limit exists for certain, but how to show that it is equal to zero then?

Answer 1

$$\lim_{r\rightarrow 0^+} r\ln\ r = \lim_{r\rightarrow 0^+} \frac{\ln\ r}{1/r}=\lim_{r\rightarrow 0^+} \frac{\frac{1}{r}}{-\frac{1}{r^2}} =0 $$

Then $$ |y\ln\ (x^2+y^2) - 0 | \leq |r \ln\ r^2|=2r|\ln\ r|\rightarrow 0 $$ where $r:=\sqrt{x^2+y^2}\rightarrow 0 $.

Answer 2

Considering $x=rcos(\theta)$ and $y=rsin(\theta)$ would lead you to $lim_{r\to 0}$$2r$sin($\theta$)$ln(r)$ that tends to zero as $r$ tends to $0$.

Answer 3

We have $$ \ln(y^2)\leq\ln(x^2+y^2)\leq\ln(1+x^2+y^2). $$ It follows that $$ |\ln(x^2+y^2)|\leq\max\{|\ln(y^2)|, \ln(1+x^2+y^2)\}, $$ and so $$ |y\ln(x^2+y^2)|\leq\max\{|y\ln(y^2)|, |y|\ln(1+x^2+y^2)\}. $$ Since $$ \lim_{(x,y)\rightarrow(0,0)}|y\ln(y^2)|=\lim_{(x,y)\rightarrow(0,0)}|y|\ln(1+x^2+y^2)=0, $$ $$ \lim_{(x,y)\rightarrow(0,0)}y\ln(x^2+y^2)=0. $$