Answer
$4$
Work Step by Step
Since, we are given that $a_{n+1}=\sqrt {8+2a_n}$ and $a_1=0$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {8+2a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=\sqrt {8+2l}$
or, $l^2-2l-8=0$
or, $l=-2 $ or, $4$
Since, we have $a_1=0$, which is $0 \gt 0$ and $a_2=\sqrt 8 \gt 0$
As per the Principal of Mathematical Induction $a_n \geq 0$, which means that $ l \geq 0$
Hence, the limit of the sequence is $4$
