Calculate limit of $(1 + x^2 + y^2)^\frac{1}{x^2 + y^2 + xy^2}$ as $(x,y) \rightarrow (0,0)$

Question

I'm trying to calculate the limit of $(1 + x^2 + y^2)^\frac{1}{x^2 + y^2 + xy^2}$ as $(x,y) \rightarrow (0,0)$.

I know that the limit is supposed to be $e$, and I can arrive at this answer if I study the univariate limit by, for instance, setting $x=t, y=0$ and letting $t \rightarrow 0$ or vice versa.

I'm not sure how to calculate it for the multivariate case. I tried using polar coordinates $(x= r \cos{\theta}, y = r \sin{\theta})$ which gets me

$$ (1 + r^2)^{\frac{1}{r^2(1 + r \cos{\theta}\sin^2{\theta})}} $$, after symplifying the expression, but I don't know how to proceed from there. I assume that I am supposed to end up with an expression similar to $(1 + n)^\frac{1}{n} \rightarrow e$ as $n \rightarrow 0$.

Answer 1

From here

$$(1 + r^2)^{ \frac{1}{r^2(1 + r \cos{\theta}\sin^2{\theta})} }=e^{\frac{\log (1 + r^2)}{r^2(1 + r \cos{\theta}\sin^2{\theta})}}\to e$$

indeed

$$\frac{\log (1 + r^2)}{r^2(1 + r \cos{\theta}\sin^2{\theta})}=\frac{\log (1 + r^2)}{r^2}\frac{1}{1 + r \cos{\theta}\sin^2{\theta}}\to1\cdot1=1$$

Answer 2

You seem quite close. The condition $(x,y) \to 0$ translates into the polar condition $r \to 0$. Applying that to the expression you have, all $\theta$ terms disappear. The rest looks quite similar to $\lim_{n\to0}(1+n)^\frac{1}{n}$.

Answer 3

I thought it might be instructive to present an approach that circumvents use of polar coordinates and relies instead on a straightforward application of the AM-GM inequality. To that end we proceed.

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To begin, the AM-GM inequality guarantees that for $x>-1$

$$\begin{align} \left|\frac{xy^2}{x^2+(1+x)y^2}\right|&\le \frac{|x|y^2}{2|x||y|\sqrt{1+x}}\\\\ &=\frac{|y|}{2\sqrt{1+x}} \end{align}$$

Hence, applying the squeeze theorem reveals

$$\lim_{(x,y)\to(0,0)}\left(\frac{xy^2}{x^2+(1+x)y^2}\right)=0\tag1$$

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Next, we can write

$$\frac{x^2+y^2}{x^2+y^2+xy^2}=1-\frac{xy^2}{x^2+(1+x)y^2}$$

Appealing to $(1)$, we have

$$\lim_{(x,y)\to(0,0)}\left(\frac{x^2+y^2}{x^2+y^2+xy^2}\right)=1\tag2$$

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Finally, equipped with $(2)$, we find that

$$\begin{align} \lim_{(x,y)\to(0,0)}\left(\left(\left(1+x^2+y^2\right)^{\frac1{x^2+y^2}}\right)^{\frac{x^2+y^2}{x^2+y^2+xy^2}}\right)&=\left(\lim_{(x,y)\to(0,0)}\left(\left(1+x^2+y^2\right)^{\frac1{x^2+y^2}}\right)^{\lim_{(x,y)\to(0,0)}\left(\frac{x^2+y^2}{x^2+y^2+xy^2}\right)}\right)\\\\ &=e^1\\\\ &=e \end{align}$$

And we are done!