Answer
$1 +\sqrt 2 $
Work Step by Step
Since, we are given that $a_{n+1}=2+\dfrac{1}{a_n}$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}2+\dfrac{1}{a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=2+\dfrac{1}{l}$
or, $l^2-2l-1=0$
or, $l=1 +\sqrt 2 $ or, $1 -\sqrt 2$
Since, we have $a_n\gt 0$ for all $n \geq 1$
Hence, the limit of the sequence is $1 +\sqrt 2 $
