An Alternate Proof to a Theorem Involving "e" 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?
 A: 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.
A: 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 $o(1)$)
$$\left(1 - \frac{{f(x)}}{x}\right)^{x}  = e^{-f(x) + o(1)}
\sim e^{-f(x)}.$$
