Squeeze theorem question for limits

$$\Bigg|\frac{5xy^2}{x^2+y^2} \Bigg|\le\Bigg|\frac{5xy^2}{y^2} \Bigg|=|5x|$$

Note that $${ x }^{ 2 }-2xy+{ y }^{ 2 }\ge 0\\ { x }^{ 2 }+{ y }^{ 2 }\ge 2xy\\ \frac { 1 }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } \le \frac { 1 }{ \left| 2xy \right| } $$ so $$ \\ \\ \\ \\ \left| \frac { 5xy^{ 2 } }{ x^{ 2 }+y^{ 2 } } \right| \le \left| \frac { 5x{ y }^{ 2 } }{ 2xy } \right| =\frac { 5 }{ 2 } \left| y \right| $$

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.