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The limit $$ \lim_{(x,y)\to(0,0)} \frac{x^{1/3}y^2}{x+y^3} $$ does exist since when we set $y=0$, $x=0$, and $x=y$ (it all equals to $0$). So now we need to evaluate the limit. The L'Hospital rule does not apply with multiple variables, so I am stuck of how to approach it. I understand that I need to somehow manipulate it, but I have no idea how. Any hints?

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    $\begingroup$ Does it? What about the limits along the paths $x = y^3$ and $x = 2y^3$? $\endgroup$ – stochasticboy321 Oct 6 '17 at 3:56
  • $\begingroup$ It doesn't exist. $\endgroup$ – IntegrateThis Oct 6 '17 at 3:56
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Are you sure that it exists? The fact that you get the same value of the limit along three different trajectories is NOT sufficient to conclude that the limit exists. The limit must be the same along all possible trajectories (approaching the given point). And you have only tried three, not all, so it's way too early to jump to any conclusions.

Hint: try the curve $x=y^3$ as the trajectory. To express the fact that $(x,y)\to(0,0)$ along this trajectory, substitute $x=y^3$ and then find the limit as you let $y\to0$ (along this curve).

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