Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a good approach? $0<|(x^5)<(x^2)^2|=|x|$ so $|x^5+y^5|<|(x^2+y^2)^2|$ so therefor 
$$0< \dfrac{|x^5+y^5|}{|(x^2+y^2)^2|}<1$$ then multiply both sides with $\ln(x^2+y^2)$. and take the $$\lim_{(x,y)\to (0,0)} \ln(x^2+y^2).$$ and since it is not $0$ but its negative infinity the limit doesn't exist by the squeeze theorem. which is not the right answer.

For all norm we have $|x|, |y|\le \|(x,y)\|$
then we get $$|x^5+ y^5|\le 2\|(x,y)\|^5$$
Since $\lim_{x\to 0}x\ln x =  0$ and $\ln(x^2+y^2) = 2\ln\|(x,y)\| $  we obtain,
$$\left|\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right| \le \lim_{\|(x,y)\|\to 0} 4\|(x,y)\|\left|\ln\|(x,y)\|\right| =0$$
