Limit of $\frac{y^4\sin(x)}{x^2+y^4}$ when $(x,y) \to (0,0)$

Question

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?

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

Answer 1

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 $v=y^2$

$$\frac{y^4x}{x^2+y^4}=\frac{uv^2}{u^2+v^2}=r\cos \theta\sin^2 \theta \to 0$$

Answer 2

Using AM-GM then we get $$\frac{x^2+y^4}{2}\geq |x|y^2$$ so $$\frac{y^4|x|}{x^2+y^4}\le \frac{y^2}{2}$$ and this tends to zero if $y$ tends to zero. We also have used that $$|\sin(x)|\le 1$$

Answer 3

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 : $|\sin x| \le |x|$