Answer
$\dfrac{1 +\sqrt 5}{2} $
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$ this implies that there exist $n$ for all natural numbers N.
Given: $a_{n+1}=\sqrt {1+a_n}$
Let $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {1+a_n} ; \\ l=\lim\limits_{n \to \infty} a_n$
Then, $l=\sqrt {1+l} \implies l^2=1+l$
and $l^2-l-1=0 \implies l=\dfrac{1 +\sqrt 5}{2} , \dfrac{1 -\sqrt 5}{2}$
As we are given that $a_n\gt 0$ for all $n \geq 1$
Thus, the limit of the sequence is : $l=\dfrac{1 +\sqrt 5}{2} $
