Let $B = \max\{|f(a)|+1, |L+1|,|L-1|\}$. Let $\epsilon =1 >0$.
Since $\lim_{x \to a}f(x)= L$, there is a number $\delta > 0$ where
if $a-\delta<x<a+ \delta$ and $x\ne a$ then $L-\epsilon <
f(x) < L + \epsilon $.
Let $r= a -\delta$ and $s= a + \delta$. Then if $x \in (r,s)$, it is not hard to show that $|f(x)|<B$.
EOP
Below is a GeoGebra visualization of this argument using mapping
diagrams applied to $f(x)= 3x-2$ and $a =3$.
You can change the values of $a$ as well as the
function $f$ to explore the arguments further with other examples
Martin Flashman, 20 Jan 2018, Created with GeoGebra