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Compute the limit (if exists) $$\lim_{(x,y)\rightarrow (0,0)}e^{\frac{-y^2}{x^4}}\sqrt[3]{y}.$$

My attempt: We can pass through polar coordinates: $$\Bigg\lvert e^{\frac{-\sin^{2}(\theta )}{\rho^{2}\cos^{4}(\theta)}}\sqrt[3]{\rho\sin (\theta)}\Bigg\rvert\leq$$ $$\leq\Bigg\lvert e^{\frac{-\sin^{2}(\theta )}{\rho^{2}\cos^{4}(\theta)}}\rho^{\frac{1}{3}}\Bigg\rvert.$$ Now, if $\theta\neq \frac{\pi}{2}$, we have that the function approach $0$ for $\rho\rightarrow 0$, and if $\theta\rightarrow \frac{\pi}{2}$ the function converges aswell (it is clear if we control the directions $y=mx$). Anyway, I'm not sure this is enough to conclude.

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  • $\begingroup$ I would argue that $e^{-x} \in (0,1]$ is always a finite value and hence you only have to check $\sqrt[3]{y}$ which goes to zero. Not sure, if this argument is 100% rigorous though. $\endgroup$ Commented Sep 4 at 8:17

2 Answers 2

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A simpler way:

Note that $$\left\vert e^{\frac{-y^2}{x^4}}\sqrt[3]{y} \right\vert\leq\vert \sqrt[3]{y}\vert $$

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A simple solution has been given already. But your question about whether your approach is correct has not been answered. You cannot merely check that the limit exists if the direction is fixed. It is possible that a function has a limit at $(0,0)$ if restricted to any line through $(0,0)$, but does not have a limit at $(0,0)$.

For example, let $f(x,y) = \cases{ 1 & if $x \ne 0$ and $x^2 = y$ \\ 0 & otherwise}$. Then the limit of $f$ when restricted to any line through $(0,0)$ is $0$, but $f$ is discontinuous at $0$.

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