## University Calculus: Early Transcendentals (3rd Edition)

We can guess here that $$\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}=\frac{1}{3}$$
$$\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}$$ (a) The graph is shown below. (b) Looking at the graph, as $x\to0$, we see that $\frac{\sqrt[3]{1+x}-1}{x}$ approaches as close as $\frac{1}{3}\approx0.333333333...$. Therefore, we can guess here that $$\lim_{x\to0}\frac{\sqrt[3]{1+x}-1}{x}=\frac{1}{3}$$