Answer
Proof given below.
Work Step by Step
Given a positive value $\epsilon$, we observe $\epsilon/2$, for which, by definition of limits, there exists an $N$ such that
$ n \gt N\displaystyle \quad \Rightarrow \quad |a_{n}-L_{1}| \lt \frac{\epsilon}{2}\quad$ and $ |a_{n}-L_{2}| \lt \displaystyle \frac{\epsilon}{2}$
Then,
$|L_{1}-L_{2}|=|(L_{1}-a_{n})+(a_{n}-L_{2})|\leq|L_{1}-a_{n}|+|a_{n}-L_{2}|$
(each of the terms in the last expression is less than $\epsilon/2$)
So, if $n\gt N.$
$|L_{1}-L_{2}| \lt \displaystyle \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
$\epsilon$ can be as small as we wish, so $|L_{1}-L_{2}| $ must be smaller than any positive number.
The only way this is true, is when $|L_{1}-L_{2}| =0$,
that is, when
$L_{1}=L_{2}.$
