#### 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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$f(x)=x^{2}.$
Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that
$ 0 < |x-0| < \delta\ \ \Rightarrow\ \ |x^{2}-0| < \epsilon$.
The conclusion of this implication...
$|x^{2}-0| < \epsilon$
$| x^{2}| < \epsilon$
$ x^{2} < \epsilon$
$|x| < \sqrt{\epsilon}$
... is equivalent to $|x| < \sqrt{\epsilon}$.
So, for $\epsilon > 0$, we take $\delta=\sqrt{\epsilon}$, for which
$ 0 < |x-0| < \delta \ \ \Rightarrow\ \ |x^{2}-0| < \epsilon$
By the definition,
$\displaystyle \lim_{x\ \rightarrow 0}x^{2}=0$