University Calculus: Early Transcendentals (3rd Edition)

$9$
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 $9$