Comment on $x$, if $x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$. The main question goes like this:
$$x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$$
Comment on $x$.
There are options given as well: 
A) $x$ is an irrational number 
B) $2<x<3$ 
C) $x=3$ 
D) None of these 
My approach:
So I try to square the term first. Pretty obvious, it's not going to get me anywhere, I think. I think $x$ is irrational, but I don't have any solid reason. Please help me out.

Let $a_{n+1}=\sqrt{6+a_n}$, $a_1=\sqrt{6}$.
We'll prove that $a_n<3$.
Since $a_1<3$, it remains to prove that $a_{n+1}<3$ for $a_n<3$.
Indeed, $a_{n+1}=\sqrt{a_n+6}<\sqrt{3+6}=3$.
Id est, $a_n<3$.
By another hand, $a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{(3-a_n)(2+a_n)}{\sqrt{a_n+6}+a_n}>0$,
which says that there is $\lim\limits_{n\rightarrow+\infty}a_n$.
Let $\lim\limits_{n\rightarrow+\infty}a_n=A$.
Hence, $\lim\limits_{n\rightarrow+\infty}a_{n+1}^2=\lim\limits_{n\rightarrow+\infty}a_n+6$ or $A^2=A+6$ or $A=3$, which gives the answer. 
