As per the definition of limits if $\underset{x\to a}{lim}f(x)=L$, then

$\mathrm{\forall}\epsilon >0\text{}\mathrm{\exists}\delta 0\text{}s.t0|x-a|\delta \text{}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}0|f(x)-L|\epsilon $

$\mathrm{\forall}\epsilon >0\text{}\mathrm{\exists}\delta 0\text{}s.t0|x-a|\delta \text{}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}0|f(x)-L|\epsilon $