## Calculus: Early Transcendentals 8th Edition

On the graph, we can see that both ratios have the same limit as $x \to 0$ $\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = \frac{1}{4}$
On the graph, we can see that both ratios have the same limit as $x \to 0$ $\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = \lim\limits_{x \to 0}\frac{e^x-1}{x^3+4x} = \frac{0}{0}$ We can apply L'Hospital's Rule. $\lim\limits_{x \to 0}\frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0}\frac{e^x}{3x^2+4} = \frac{1}{0+4} = \frac{1}{4}$ Therefore: $\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = \frac{1}{4}$