Is ma solution correct to this equation? $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3$$
My take: $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3\implies {x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=9$$
Thus: $$x+3=9\implies x=6$$
Did I do it right?

One way to compute this limit is the following:
Let $x_1 = x $, and $x_n = \sqrt{x + x_n}$, for $n \geq 2$.  Then prove that the limit of the $x_n$'s exists by showing that the sequence is monotonic and bounded.
Then, we have that $L =  \sqrt{x + L}$ but $L =3$ so you get that $x = 6$. 
