How to find $\lim_{(x,y)\rightarrow(0,0)} \frac{x^2\sin^2(y)}{x^2+3y^2}$ Find the following $$\lim_{(x,y)\rightarrow(0,0)}  \frac{x^2\sin^2(y)}{x^2+3y^2}$$
How to solve this limit? I have tried to use polar system, like sub everything with $x=rcos\theta$ and $y=\sin\theta$, but r cancel each other.


$$0 \leq \frac{x^2\sin^2(y)}{x^2+3y^2}\leq \frac{x^2\sin^2(y)}{3y^2} =\frac{1}{3}x^2 \frac{\sin{y}}{y} \frac{\sin{y}}{y} \longrightarrow 0 \times 1 \times 1=0$$ as 
   $(x,y) \to (0,0)$

Thus from squeeze theorem the limit is $0$
