If

s.t.

we get

Look you must start with Suppose that $C>0$. Everything in the proof depends upon that positive number.

The expression $|L-a_n|$ is the distance from $a_n$ to $L$.

So expression $|L-a_n|<C$ means that is the distance from $a_n$ to $L$ is less than $C$.

Now $|L-a_n|<C$ is the open interval $(L-C,L+C)$ so that $a_n\in(L-C,L+C)$.

This also means that you can find a positive integer $N_1$ (it depends upon $C$) having the property that picking any integer, $j$, greater than $N_1$ then $a_j\in(L-C,L+C)$

**or** $|a_j-L|<C$

So roughly speaking all that means is Given a positive error, C, we can find a place in the sequence where from there on the terms are close to L.