Answer
$5$
Work Step by Step
Since, we are given that $a_{n+1}=\sqrt {5a_n}$ and $a_1=5$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {5a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=5l$
or, $l^2-5l=0$
or, $l=0 $ or, $5$
Since, we have $a_1=5$, which is $5 \gt 1$ and $a_2=\sqrt {(5)(5)} \gt 1$
As per the Principal of Mathematical Induction $a_n \geq 1$, which means that there exists n for all natural numbers N.
Hence, the limit of the sequence is $5$
