Asymptotic relations at infinity I am attempting to show that

If $f(x) - g(x) \ll 1,\, x \to \infty$, then $e^{f(x)}\sim e^{g(x)}, \,x\to \infty$

From the first line, I am able to show that
$$ \lim_{x\to \infty} \frac{f(x) - g(x)}{1} = 0$$
from which it is clear that
$$ \lim_{x\to \infty}\left[f(x) - g(x)\right] = 0$$
However, I am a little stuck at this point. Where do exponentials come in?
Any help/hints much appreciated.

As you've shown 
$$f(x)-g(x)\to 0,\quad\text{as}\, x\to\infty$$
So 
$$\frac{e^{f(x)}}{e^{g(x)}}=e^{f(x)-g(x)}\to 1$$
All you need is
$$e^v\to 1,\quad\text{as}\,v\to 0$$
