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In a paper, it was claimed that $\lim_{x \to \infty} (1-\frac{f(x)}{x})^x \sim e^{-f(x)}$ when $\frac{(f(x))^2}{x}$ is $o(1)$.

I proved the claim in the following way; however, I'm seeking a simpler proof.

Define $g(x) = \frac{(1-f(x)/x)^x}{x}$ and $h(x) = \frac{e^{-x}}{x}$. To prove the theorem, we must show that $\lim_{x \to \infty} \frac{g(x)}{h(f(x))} = 1$ when $\frac{(f(x))^2}{x}$ is $o(1)$. To this end, we use the binary expansion for $g(x)$, and Taylor series for $h(x)$:

$\lim_{x \to \infty} g(x) = \lim_{x \to \infty} \frac{(x-f(x))^x}{x^{x+1}} = \lim_{x \to \infty} \frac{x^x - \binom{x}{1}x^{x-1}f(x) + \binom{x}{2}x^{x-2}(f(x))^2 - \cdots}{x^{x+1}} = \lim_{x \to \infty} \frac{1-f(x)}{x}$

(The last identity follows from the fact that $\frac{(f(x))^2}{x}$ is $o(1)$; that is $\lim_{x \to \infty} \frac{(f(x))^2}{x} = 0$.)

Now, since $h(x) \sim \frac{1 - x + x^2/2 - \cdots}{x}$, we have $h(f(x)) \sim \frac{1 - f(x) + (f(x))^2/2 - \cdots}{x}$, and $\lim_{x \to \infty} h(f(x)) = \lim_{x \to \infty} \frac{1-f(x)}{x}$. (Again, the last identity follows from the fact that $\frac{(f(x))^2}{x}$ is $o(1)$.)

Combining the two limits, we see that $\lim_{x \to \infty} \frac{g(x)}{h(f(x))} = 1$.

As I said, this proof is long, and to me, it is not appealing.

Does anyone know a better proof?

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I think using the expansion of the $\ln$ is more natural than the binomial expansion. Not that that shortens the proof a lot. – Raskolnikov Mar 29 '11 at 21:45
up vote 13 down vote accepted

Let'sconsider $$\lim_{x\to\infty} \left(1 - \frac{{f(x)}}{x}\right)^{x} = \lim_{x\to\infty} \exp\left[x \ln \left(1 - \frac{f(x)}{x}\right)\right].$$ Expanding the logarithm, we obtain $$x \ln \left(1 - \frac{f(x)}{x}\right) = x \left[ - \frac{f(x)}{x} + O\left(\frac{f(x)}{x}\right)^2\right] = -f(x) + O\left(\frac{f(x)^2}{x}\right).$$ So we obtain (using the fact that $f(x)^2/x$ is $o(1)$) $$\left(1 - \frac{{f(x)}}{x}\right)^{x} = e^{-f(x) + o(1)} \sim e^{-f(x)}.$$

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+1, but the end should be revised. – Shai Covo Mar 29 '11 at 22:33
@Shai Covo: I will try to fix that; I forgot that $f(x)$ depends on $x$... ;-) – Fabian Mar 29 '11 at 23:03

It might be useful to consider $(1 - \frac{{f(x)}}{x})^{(x/f(x))f(x)}$.

EDIT: This is a comment (in response to the OP's comment below).

First of all, note that without loss of generality we can suppose that $f(x) \neq 0$. Now, at first glance, it seemed that writing $(1 - \frac{{f(x)}}{x})^x $ as $(1 - \frac{{f(x)}}{x})^{(x/f(x))f(x)}$ might be useful, since $(1 - \frac{{f(x)}}{x})^{x/f(x)} \to e^{-1}$. Indeed, note that since $\frac{f^2(x)}{x}$ is $o(1)$, $f(x)/\sqrt{x} \to 0$, hence also $f(x)/x \to 0$. However, this approach only gives intuition for the desired asymptotic equality, and beyond that is apparently not useful in our context. Fabian's approach seems the best to me.

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Too many upvotes for this hasty suggestion... – Shai Covo Mar 29 '11 at 22:01
Should I balance them out?? ;) – The Chaz 2.0 Mar 30 '11 at 1:06
@The Chaz: I will try to improve the answer... – Shai Covo Mar 30 '11 at 1:35
@Shai Covo: I was joking! You are very helpful. Have a good day! – The Chaz 2.0 Mar 30 '11 at 2:22
@The Chaz: Thanks! – Shai Covo Mar 30 '11 at 3:01

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