Answer
$8$
Work Step by Step
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 $l=8$.
