Answer
$9$
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$ this implies that there exist $n$ for all natural numbers N.
Given: $a_{n+1}=12-\sqrt {a_n}$
Suppose $l=\lim\limits_{n \to \infty} a_n$; $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}12-\sqrt {a_n}$
Then $l=12-\sqrt l$
or, $l^2-25l+144=0$
Thus, $(l-9)(l-16)=0 \implies l=9 , 16$
Since, we have $12-\sqrt {a_n} \lt 12$
Thus, the limit of the sequence is $l=9$
