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