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I am tasked with evaluating $$\lim_{(x, y) \to (0, 0)} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}.$$

We can rewrite the expression it as follows:

$$ e^{\ln(1+x^2+y^2)\cdot\frac{1}{x^2+y^2+x y^2}} $$

Now, my first guess was that this tends to $1$. But that is apparently incorrect, it tends towards $e$, which one can see just by graphing it. That is also what the book says.

Now, we have the "standard limit":

$$\lim_{x \to 0^+} \frac{\ln(1+x)}x = 1$$

So I can kinda see how we can get the power to tend toward $1$ so we get $e^1$, but the expression is not exactly the same, we have $xy^2$ tacked on to the end of the denominator of the fraction. Furthermore, $xy^2$ doesn't have to be positive.

I tried using the substitution $x=\frac{1}{y^2}$, that gives us an expression that tends towards infinity both at $0$ and infinity and is never zero.

Can anyone point out to me what I'm missing. It feels like it should be super-obvious.

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4 Answers 4

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Your rewrite is a step in the right direction. Next, use polar coordinates. The limit becomes: $$\lim_{\rho \to 0} \exp\left(\frac{\ln(1 + \rho^2)}{\rho^2(1 + \rho\sin^2\theta\cos\theta)}\right)$$

Using the fact that $\ln(1 + t) \sim_0 t$, and that $\sin^2\theta\cos\theta$ is bounded, you can easily finish.

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  • $\begingroup$ Ahhrgh.. of course, polar coordinates. Thanks! I should've realized the authors intentions from $x^2+y^2$. By the way what does the tilde with a subscript zero mean? Never seen that before. $\endgroup$
    – Skurmedel
    Dec 17, 2015 at 0:16
  • $\begingroup$ The key idea is polar coordinates, they're very useful for two-dimensional limits at the origin. $\endgroup$
    – JacksonFitzsimmons
    Dec 17, 2015 at 0:58
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    $\begingroup$ @Skurmedel Yes, polar coordinates are very useful because if you manage to show that the limit does not depend on $\theta$, you have a limit in one variable, while still approaching the limit point from all directions. And as for the notation $\sim_0$, it's just a shortand for "$\ln(1 + t) \sim t$ as $t \to 0$", or, equivalently, "$\ln(1 + t) = t + o(t)$ as $t \to 0$". This is all equivalent to the standard limit you stated in the question. $\endgroup$
    – rubik
    Dec 17, 2015 at 9:47
  • $\begingroup$ @rubik: Yeah I see, very neat :) Thanks for the explanation. $\endgroup$
    – Skurmedel
    Dec 17, 2015 at 20:43
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The given limit exists if and only if the limit $$ \lim_{(x,y)\to(0,0)}\frac{\ln(1+x^2+y^2)}{x^2+y^2+xy^2} $$ exists. Pass to polar coordinates: $x=r\cos\varphi$, $y=r\sin\varphi$, so the limit becomes $$ \lim_{r\to0}\frac{\ln(1+r^2)}{r^2(1+r\cos\varphi\sin^2\varphi)}= \lim_{r\to0}\frac{\ln(1+r^2)}{r^2} \lim_{r\to0}\frac{1}{1+r\cos\varphi\sin^2\varphi}=1 $$

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  • $\begingroup$ Thank you. and nice explanation. I missed something obvious yet again :) $\endgroup$
    – Skurmedel
    Dec 17, 2015 at 0:17
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Let $x^2+y^2=r^2 .$ We have $$|x|<a\implies |x y^2|\leq a y^2\leq a(x^2+y^2)=a r^2 .$$ So let $xy^2=r^2(1+b)$ where $b\to 0$ as $r\to 0 .$ Observe that this implies that $x^2+y^2+xy^2>0$ for sufficiently small positive $r.$ So, for all sufficiently small $r>0$ we have $$\log (1+x^2+y^2)^{1/(x^2+y^2+x y^2)}=$$ $$ =\frac {1} {1+b} \frac {1}{r^2}\log (1+r^2)$$ which tends to $1$ as $r\to 0.$

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If we write $u = x^2+y^2,$ the expression equals

$$[(1+u)^{1/u}]^{u/(u+xy^2)} = [(1+u)^{1/u}]^{1/(1+(xy^2/u))} .$$

As $u\to 0,$ the expression inside the brackets $\to e.$ It's a simple argument to show $xy^2/u \to 0.$ Thus the exponent outside the brackets $\to 1,$ and the desired limit is $e^1=e.$

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