Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 91

$8$

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