Answer
Proof given below.
Work Step by Step
Let $\{a_{n}\}$ converge to $L$.
Let $f$ be a function continuous at L, defined for all $a_{n}.$
Select any $\epsilon\gt 0.$
Since f is continuous at L, there exists a $\delta\gt 0$ such that
$|x-L|\lt \delta \Rightarrow |f(x)-f(L)|\lt \epsilon.\qquad(*)$.
For the $\delta\gt 0$ from above, since $a_{n}\rightarrow L$, there exists an N such that
$ n\gt N\Rightarrow|a_{n}-L|\lt \delta$
Applying $(*)$, it follows that
$ n\gt N\Rightarrow |f(a_{n})-f(L)|\lt \epsilon$.
This means that $f(a_{n})\rightarrow f(L)$
