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