Answer
$\lim\limits_{x \to 0}\frac{f(x)}{e^x-1} = 1$
Work Step by Step
$\lim\limits_{x \to 0}\frac{f(x)}{e^x-1} = \lim\limits_{x \to 0}\frac{x}{e^x-1} = \frac{0}{0}$
We can use L'Hospital's Rule:
$\lim\limits_{x \to 0}\frac{f'(x)}{\frac{d}{dx}(e^x-1)} = \lim\limits_{x \to 0}\frac{1}{e^x} = \frac{1}{1} = 1$
$\lim\limits_{x \to 0}\frac{f(x)}{e^x-1} = 1$