Answer
$\dfrac{1 +\sqrt 5}{2} $
Work Step by Step
Since, we are given that $a_{n+1}=\sqrt {1+a_n}$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {1+a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=\sqrt {1+l}$
or, $l^2-l-1=0$
or, $l=\dfrac{1 +\sqrt 5}{2} , \dfrac{1 -\sqrt 5}{2}$
Since, we have $a_n\gt 0$ for all $n \geq 1$
Hence, the limit of the sequence is $\dfrac{1 +\sqrt 5}{2} $
