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.

Hint. Note that by letting $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, we have that as $(x,y)\to(0,0)$ then $\rho\to 0$ and
$$0\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right|\leq
\frac{(|\rho\cos\theta|^5+|\rho\sin\theta|^5)|\ln(\rho^2)|}{\rho^4}
\leq \frac{\rho^5(1+1)|\ln(\rho^2)|}{\rho^4}={4\rho|\ln(\rho)|}.$$
Can you take it from here? 
