Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 311: 7

$\lim\limits_{x \to 0}\frac{f(x)}{e^x-1} = 1$

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