Proof of CCD.LB:
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  $a =3$.

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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