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?