#### Answer

please see step-by-step

#### Work Step by Step

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.