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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

enter image description here

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  • $\begingroup$ Yes, sure, you've proved. $\endgroup$ – Mher Mar 8 '14 at 12:54
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Yes, if you find just one curve (or even a sequence) that converges towards the given point but the function values along it don't converge, then you have proved that the function at large doesn't have a limit there.

(Exercise: Prove this general fact from the $\varepsilon$-$\delta$ definition!)

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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!

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