## Thomas' Calculus 13th Edition

$1 +\sqrt 2$
According to the Principle of Mathematical Induction $a_n \geq 0$ this implies that there exist $n$ for all natural numbers N. Given: $a_{n+1}=2+(\dfrac{1}{a_n})$ Suppose $l=\lim\limits_{n \to \infty} a_n ;\\l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}2+(\dfrac{1}{a_n})$ Then, $l=2+\dfrac{1}{l}$ or, $l^2-2l-1=0 \implies l=1 +\sqrt 2$ or, $1 -\sqrt 2$ As we are given that $a_n\gt 0$ for all $n \geq 1$ Thus, the limit of the sequence is $l=1 +\sqrt 2$