## Thomas' Calculus 13th Edition

$8$
Consider $a_{n+1}=\dfrac{72}{1+a_n}$ and $a_1=2$ suppose $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\dfrac{72}{1+a_n}$ Thus, $l=\dfrac{72}{1+l}$ or, $l^2+l-72=0 \implies (l+9)(l-8)=0 \implies l=-9$ or, $8$ As we are given that $a_1=2$; and $2 \gt 0$ Hence, the limit of the sequence is $l=8$.