Showing the limit exist by using sequences

$$a_1=1,a_{n+1}=\sqrt{6+a_n}, n \in \mathbb{N}$$

Show that the limit $\lim\limits_{n \to +\infty} a_n$ exists and find it.

I know to prove that the sequence is bounded and monotonous but I still don't know how to find the limit.

One can show by induction that $a_n \leqslant 3$ and monotonically increasing, which means that the limit exists. Suppose the limit is $L$. Then,
$$L = \sqrt{6+L} \Rightarrow L^2 - L - 6=0 \Rightarrow L = 3$$
So, $\lim\limits_{n \to +\infty} a_n = L = 3$.
• @bopele At the limit $a_{n+1} = a_n = L$. Solving the equation $L^2 - L - 6 = 0$ gives you three. Knowing that show $a_n \leq 3$ by induction. – Χpẘ Mar 31 '17 at 16:43