Limit of $\lim_{(x,y)\rightarrow (0,0)}\ xy\log(\lvert x\rvert+\lvert y\rvert)$ I need help to understand how we compute this kind of limit:
$\lim_{(x,y)\rightarrow (0,0)}\ xy\log(\lvert x\rvert+\lvert y\rvert)$
I think we can use the squeeze theorem but I don't know how to bound the function, so I can use the theorem. If I suppose $0 \lt \sqrt{x^2+y^2} \lt 1$ then, but I'm struggle..  
$ 0 \le \lvert f(x,y)\rvert = \lvert xy\log(\lvert x\rvert+\lvert y\rvert)\rvert$
Thanks in advance for the help.
 A: Use the AM-GM inequality:
$$
0\leq 2\sqrt{\lvert xy\rvert} \leq |x|+|y| \tag{1}
$$
from which
$$
0\leq 2\sqrt{\lvert xy\rvert}|\log(|x|+|y|)| \leq (|x|+|y|)|\log(|x|+|y|)|\tag{2}
$$
Setting $t(x,y)\stackrel{\rm def}{=} |x|+|y|$, we have $\lim_{(x,y)\to(0,0)} t(x,y)=0$ (why?) and therefore since $\lim_{t\to 0}t\log t = 0$ we get by the Squeeze theorem
$$
\lim_{(x,y)\to(0,0)}\sqrt{\lvert xy\rvert}|\log(|x|+|y|)| = 0\tag{3}
$$
and also 
$$\lim_{(x,y)\to(0,0)}\sqrt{\lvert xy\rvert} = 0 \tag{4}$$
also by (1) and the Squeeze theorem. 
Can you conclude?
A: HINT
We have that
$$xy\log(\lvert x\rvert+\lvert y\rvert)=\frac{xy}{|x|+|y|}\cdot (|x|+|y|)\log(\lvert x\rvert+\lvert y\rvert)$$
then recall that as $u\to 0$


*

*$u\log u \to 0$


and by polar coordinates


*

*$\frac{xy}{|x|+|y|}=r\cdot\frac{\cos \theta \sin \theta}{|\cos \theta|+|\sin \theta|}$


with


*

*$|\cos \theta|+|\sin \theta|\ge m>0$


Or as an alternative directly by polar coordinates
$$xy\log(\lvert x\rvert+\lvert y\rvert)=r\cos \theta \sin \theta\cdot \left(r\log r+r\log(|\cos \theta|+|\sin \theta|)\right)$$
