TITLE: Squeeze theorem question for limits
QUESTION [7 upvotes]: How can I find the following limit using the Squeeze theorem?
$$\lim_{x,y\to 0} \frac{5xy^2}{x^2+y^2}$$
I am not sure what to use for the lower and upper bounds.

REPLY [2 votes]: You could try converting into polar co-ordinates
which gives 
$$\lim_{r \to 0}\frac{5r^3\cos(\theta)\sin^2{\theta}}{r^2(\cos^2{(\theta)}+\sin^2(\theta))}=\lim_{r \to 0}5r\cos(\theta)\sin^2{\theta}=0$$
$$|\sin(\theta)| \leq1$$ $$|\cos(\theta)| \leq1$$
$$|\sin^2(\theta)\cos(\theta)| \leq1 $$
Then 
$$|5r\sin^2(\theta)\cos(\theta)| \leq 5r$$ 
Since $\lim_{r \to 0}(-5r)=\lim_{r \to 0}(5r)=0 $
Thus $r$ is bounded which means our step was justified even though $\theta $
is arbitrary
By Squeeze theorem
$$\lim_{r \to 0}5r\cos(\theta)\sin^2{\theta}=0$$
EDIT : AS mentioned by Yves Daoust,
$$|\sin^2(\theta)\cos(\theta)|\leq \frac{2\sqrt3}{9}$$
which can be verified by taking derivatives and also givs a tighter bound.