Any idea on how to find a upper bound for a limit in R^2? I'm working on this problem of continuity in $\mathbb{R}^2$ the statement is the following: 
Prove by definition that 
$$ \lim_{ (x,y) \to (0,0) } \frac{y^2}{3+\sin^2(x^2+y^2)+y^4}=0 $$
Given $\epsilon >0$, we have that 
$$| \frac{y^2}{3+\sin^2(x^2+y^2)+y^4} | =  y^2 \frac{1}{3+\sin^2(x^2+y^2)+y^4}$$
If we know that 
$ \frac{1}{3+\sin^2(x^2+y^2)+y^4} \leq  \frac{1}{\sin^2(x^2+y^2)} $ then we end with 
$$ y^2 \frac{1}{3+\sin^2(x^2+y^2)+y^4} \leq y^2 
\frac{1}{\sin^2(x^2+y^2)}$$
and even more, 
$$y^2 
\frac{1}{\sin^2(x^2+y^2)} \leq  \frac{x^2+y^2}{\sin^2(x^2+y^2)} $$
I don't see any way to get rid out the $\frac{1}{\sin^2(x^2+y^2)}$ to conclude the $\delta$ since I can't say that $\frac{1}{\sin^2(x^2+y^2)} \leq 1$ because is the contrary, in fact  $\frac{1}{\sin^2(x^2+y^2)} \geq 1$. Any ideas?
Thanks in advance. 

$\frac{y^2}{3+\sin^2(x^2+y^2)+y^4}\leq \frac{y^2}{3}\rightarrow 0$
