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