## Calculus 8th Edition

By definition , $\displaystyle \lim_{x\rightarrow 0}\sqrt[3]{x}=0$ is true if for any $\epsilon > 0$, there is a $\delta > 0$ such that if $0 < |x-0| < \delta$, then $|\sqrt[3]{x}-0| < \epsilon$. analyzing the last expression, $|\sqrt[3]{x}-0|=|\sqrt[3]{x} | < \epsilon$ $|x|=|\sqrt[3]{x}|^{3} < \epsilon^{3}$ So, for any given $\epsilon,$ if we choose $\delta=\epsilon^{3}$,we will have $0 < |x-0| < \delta\ \ \Rightarrow\ \ |\sqrt[3]{x}-0| < \epsilon$. Thus, $\displaystyle \lim_{x\rightarrow 0}\sqrt[3]{x}=0$ by the definition of a limit.