Answer
$1 +\sqrt 2 $
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$.
Given: $a_{n+1}=2+\dfrac{1}{a_n}$
Suppose $l=\lim\limits_{n \to \infty} a_n ;\\l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\left(2+\dfrac{1}{a_n}\right)$
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 $l=1 +\sqrt 2$.
