Nice sequence bound Consider the sequence so defined by induction:
$x_0 = 1$
$x_{n+1} = \sqrt{1+x_n}$
Prove that it is bounded and find the limit towards +infinity.
To me it seems to approach phi but don't know how to prove it.
 A: Boundedness is an easy induction. Note that $x_0 = 1 < 2$. Suppose $0 < x_n < 2$. Then $x_n < \sqrt{1 + x_n} < \sqrt{3} < 2$. It follows that $x_n < 2$ for all $n$, so by the monotone convergence theorem $x_n$ converges. As for the limit, note that if the sequence $a_n = f^n(x)$ converges to $L$, where $f$ is continuous on an interval containing $L$ and $f^n$ is the $n$th composition of $f$ with itself, we must have $f(L) = L$. What can you conclude about the limit from this?
A: See that the function defined in $\Bbb R_{+}^{*}$
$$f(x)=\sqrt{1+x} \rightarrow f'(x)=\frac{1}{2\sqrt{1+x}}>0$$
So $f(x)$ is monotonic increasing and then the sequence $x_{n+1}=\sqrt{1+x_n}$ is also monotonic increasing.
Once we know that $x_{n+1}>x_n$ then
$$x_{n+1}^2>x_n^2 \rightarrow x^2_n-x_n-1<0 \rightarrow \frac{1-\sqrt{5}}{2}<x_n< \frac{1+\sqrt{5}}{2}$$
So the sequence is bounded and increasing then it converge to $L$.
Doing $n \rightarrow \infty$ at $x_{n+1}^2=x_n+1$ we get:
$$L^2=L+1 \rightarrow L=\frac{1\pm\sqrt{5}}{2}$$
But $L>0$ then:
$$L=\frac{1+\sqrt{5}}{2}$$ 
