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?
$$\frac{y^4\sin(x)}{x^2+y^4}$$

We have that
$$\frac{y^4\sin x}{x^2+y^4}=\frac{\sin x}{x}\frac{y^4x}{x^2+y^4}$$
then recall that $\frac{\sin x}{x}\to 1$.
To show that $\frac{y^4x}{x^2+y^4} \to 0$ we can use inequalities or as an alternative by $u=x$ and $v=y^2$
$$\frac{y^4x}{x^2+y^4}=\frac{uv^2}{u^2+v^2}=r\cos \theta\sin^2 \theta \to 0$$
