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.
 A: I think it means
$$0< \frac{x^2\sin^2y}{x^2+3y^2}< \frac{x^2y^2}{x^2+3y^2}<x^2$$
A: 
$$0 \leq \frac{x^2\sin^2(y)}{x^2+3y^2}\leq \frac{x^2\sin^2(y)}{3y^2} =\frac{1}{3}x^2 \frac{\sin{y}}{y} \frac{\sin{y}}{y} \longrightarrow 0 \times 1 \times 1=0$$ as 
   $(x,y) \to (0,0)$

Thus from squeeze theorem the limit is $0$
A: You can use polar as follow
$$\frac{x^2 \sin(y)}{x^2+ 3y^2}=\frac{r^2 \cos^2(\theta)\sin(r \sin(\theta)}{r^2 \cos^2(\theta)+ 3r^2 \sin^2(\theta)}=\frac{ \cos^2(\theta)\sin(r \sin(\theta)}{\cos^2(\theta)+ 3 \sin^2(\theta)}\to 0$$
Since $\forall \theta $
$$\cos^2(\theta)+ 3 \sin^2(\theta) \geq m >0$$
A: The way it is written, $\frac{x^2 sin(y)}{x^2+ 3x^2}= \frac{x^2 sin(y)}{4x^2}$ the $x^2$ terms cancel, leaving $\frac{sin(y)}{4}$ which goes to 0 as y goes to 0.
  Did you mean $\frac{x^2 sin(y)}{x^2+ 3y^2}$?  In that case, replacing x by $r cos(\theta)$ and y by $r sin(\theta)$ gives $\frac{r^2 cos^2(\theta)sin(r sin(\theta)}{r^2 cos^2(\theta)+ 3r^2 sin^2(\theta)}$.  You are correct that the $r^2$ terms cancel, leaving $\frac{cos^2(\theta)sin(r sin(\theta)}{cos^2(\theta)+ 3sin^2(\theta)}$.  Since that depends on "$\theta$", there is no limit.
