# Limit as $(x,y)$ approaches $(0,0)$ of $(1+x^2+y^2)^{\frac{1}{x^2+y^2+xy^2}}$

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?

• I think Polar coordinates is not a good idea for this problem. – Pink Panther Jul 6 at 11:56
• Polar coordinates are a nice approach, but you should consider $\ln{f}$ instead. – Mindlack Jul 6 at 11:56
• Yes, Polar coordinates are a good idea, and you can easily compute the limit and find that it is equal to $e$. – uniquesolution Jul 6 at 11:59

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 $$t$$).
• Just use the definition of $\exp$ and $\log$: for $x>0$, we have that $x^y=\exp(\log(x^y))=\exp(y\log(x))$. – Robert Z Jul 6 at 12:48
• @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. – Amit Zach Jul 10 at 10:57