Answer
$5$
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$.
Given: $a_{n+1}=\sqrt {5a_n}$ and $a_1=5$
Consider $l=\lim\limits_{n \to \infty} a_n$ ;$l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {5a_n}$
Then, $l^2=5l$
so, $l^2-5l=0$
or, $l(l-5)=0 \implies l=0 $ or, $5$
As we are given that $a_1=5$ and $5 \gt 1$ and $a_2=\sqrt {(5)(5)} \gt 1$.
Thus, the limit of the sequence is $l=5$.