I need to prove this using $\epsilon$ $\delta$ $$\lim\limits_{(x,y) \to (0,0)} \frac{(x-1)^2\sin(x^2)y}{x^2+y^4} = 0$$
I know that $|x| < \delta $ and $|y| < \delta $, so then
$|\frac{(x-1)^2sin(x^2)y}{x^2+y^4}-0|$ $\lt$ $\frac{(x-1)^2|x||y|}{x^2+y^4}$ $\lt$ $\frac{(x-1)^2\delta^2}{\delta^2+\delta^4}$ $\lt$ $\frac{(x-1)^2}{1+\delta^2}$
And I have no idea how to continue from this. Any help would be appreciated!
$|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| \leq (x-1)^{2} \frac {x^{2}|y|} {x^{2}}$ since $|\sin (x^{2})| \leq x^{2}$. Hence $|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| <\epsilon$ if $|x|<1$ and $|y|<\frac {\epsilon } 4 $