<s>[INST] Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$ 
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.


Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!
 [/INST] You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then
$$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$
which completes our inductive argument and we conclude that $x_n$ converges.
Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$
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