Answer
$4$
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$ this means that $ l \geq 0$.
Given: $a_{n+1}=\sqrt {8+2a_n}$ and $a_1=0$
Consider $l=\lim\limits_{n \to \infty} a_n$ and $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {8+2a_n}$ ; $l=\lim\limits_{n \to \infty} a_n$
Thus, $l=\sqrt {8+2l}$
$l^2-2l-8=0 \implies (l+2)(l-4)=0$
so, $l=-2 $ or, $4$
Thus, the limit of the sequence is $l=4$.
