Answer
please see step-by-step
Work Step by Step
$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid:
$($if $ 0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$
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$f(x)=x^{3}.$
Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that
$ 0 < |x-0| < \delta\ \ \Rightarrow\ \ |x^{3}-0| < \epsilon$.
The conclusion of this implication...
$|x^{3}-0| < \epsilon$
$| x^{3}| < \epsilon$
$|x|^{3} < \epsilon$
$|x| < \sqrt[3]{\epsilon}$
... is equivalent to $|x| < \sqrt[3]{\epsilon}$.
So, for $\epsilon > 0$, we take $\delta=\sqrt[3]{\epsilon}$, for which
$ 0 < |x-0| < \delta \ \Rightarrow \ \ |x^{3}-0| < \epsilon$
By the definition,
$\displaystyle \lim_{x\rightarrow 0}x^{3}=0$
