# Limit of $x^3\sin y/(x^6+2y^3)$ as $(x,y)\to (0,0)$

I was asked to calculate the limit: $$\lim_{(x,y)\to (0,0)} \frac{x^3 \sin y}{x^6+2y^3}$$

I believe it has no limit so I tried to place some other functions that goes to $(0,0)$ and prove they don't goes to $0$ (I found some functions that does go to zero).

I've tried with $y=x^2$ and I find out it goes to infinity, is it good enough?

I've attached an Image explaining it better.

Thanks

• Yes, sure, you've proved. – Mher Mar 8 '14 at 12:54

(Exercise: Prove this general fact from the $\varepsilon$-$\delta$ definition!)
To be really picky, the limit as you wrote it after substituting does not exist since $x$ can approach zero from the positive or from the negative side, yielding $\pm \infty$. However you can correctly deduce that the multivariate limit does not exist either!