Limit of $\frac{y^4\sin(x)}{x^2+y^4}$ when $(x,y) \to (0,0)$ I have to find the following limit when $x,y$ tend towards $0$. I think the limit doesn't exist (thanks to Wolfram Alpha), but all I can find no matter what path I use (I've tried $y=x², x=y^2, x=0, y=0$) is that the limit equals $0$ (because of the $sin(0)$). Any tips?
$$\frac{y^4\sin(x)}{x^2+y^4}$$

Let $|x|,|y| <1$,  $x,y$ real.
$0\le \left | \dfrac{y^4 \sin x }{x^2+y^4} \right | \le $
$\dfrac{y^4|x|}{x^4 + y^4} \le$
$\dfrac{(y^4+x^4)|x|}{x^4+ y^4}= |x| \le \sqrt{x^2+y^2}.$
Choose $\delta =\epsilon.$
Used : $|\sin x| \le |x|$ 
