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I have the function $$f(x,y)=(1+x^2+y^2)^{\frac{1}{x^2+y^2+xy^2}}$$ and I want to evaluate the limit as $(x,y)$ approaches zero.

I have started thinking of a solution but get stuck. Taking the direct limit is not possible since that would make an undefined function. Approaching $(0,0)$ from different lines e.g $y=x$ and $y=0$ both gives hints that the limit could be $e$ but that does not really show anything. I tried switching to polar coordinates which gives me $$(1+r^2)^{\frac{1}{r^2+r^3\cos(t)\sin^2(t)}}$$ Was it a good idea to switch to polar coordinates, can it be solved continuing with this approach, or could the function $f$ maybe be simplified and solved in a different way?

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  • $\begingroup$ I think Polar coordinates is not a good idea for this problem. $\endgroup$ – Pink Panther Jul 6 at 11:56
  • $\begingroup$ Polar coordinates are a nice approach, but you should consider $\ln{f}$ instead. $\endgroup$ – Mindlack Jul 6 at 11:56
  • $\begingroup$ Yes, Polar coordinates are a good idea, and you can easily compute the limit and find that it is equal to $e$. $\endgroup$ – uniquesolution Jul 6 at 11:59
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You are on the right track. Just note that, as $r\to 0$, $$(1+r^2)^{\frac{1}{r^2+r^3\cos(t)\sin^2(t)}}=\exp\left(\frac{\overbrace{\frac{\log(1+r^2)}{r^2}}^{\to 1}}{1+\underbrace{r\cos(t)\sin^2(t)}_{\to 0}}\right)\to e$$ where we used the fact that, $\log(1+r^2)=r^2+o(r^2)$ and $|r\cos(t)\sin^2(t)|\leq r$ (the bound is independent of $t$).

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  • $\begingroup$ Thank you, how did you get to the equation with exp involved? I may be missing something basic but its been a while with math for me $\endgroup$ – F Wi Jul 6 at 12:45
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    $\begingroup$ Just use the definition of $\exp$ and $\log$: for $x>0$, we have that $x^y=\exp(\log(x^y))=\exp(y\log(x))$. $\endgroup$ – Robert Z Jul 6 at 12:48
  • $\begingroup$ Got it. When calculating the limit after rewriting the function do you then use the chain rule ? $\endgroup$ – F Wi Jul 6 at 13:14
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    $\begingroup$ Yes, composition of limits. $\endgroup$ – Robert Z Jul 6 at 14:14
  • $\begingroup$ @RobertZ Doesn't your solution only prove that the limit of $f$ along straight lines is $e$? What about other sorts of curves? For example, considering $\frac{x^2y}{1+x^4y^2}$, this function's limit is $0$ if you use polar coordinates, but it doesn't have a limit if you compute it along a parabola, for instance. $\endgroup$ – Amit Zach Jul 10 at 10:57

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