Calculus 8th Edition

$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid: $($if $0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$ ------------- $f(x)=|x|$ 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, $\displaystyle \lim_{x\rightarrow 0}|x|=0$