Answer
$9$
Work Step by Step
Since, we are given that $a_{n+1}=12-\sqrt {a_n}$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}12-\sqrt {a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=12-\sqrt l$
or, $l^2-25l+144=0$
or, $l=9 $ or, $16$
Since, we have $12-\sqrt {a_n} \lt 12$
Hence, the limit of the sequence is $9$
