If $x_1=\sqrt 2$ and $x_{n+1}=(\sqrt2)^{x_n}$ then sequence $x_n$ converges to 2. If $x_1=\sqrt 2$ and $x_{n+1}=(\sqrt2)^{x_n}$ then show that  sequence $x_n$ converges to 2.
I know this sequence is monotonically increasing. But how to prove it converges to 2? 
The sequence is bounded above is already answered here. And I know that$\sqrt 2 ^l=l$ has only solution 2. But how to prove formally? Thanks.
 A: Here is a formal way considering $f(x) = \left( \sqrt{2}\right)^x$ and using MVP:


*

*On $[1,2]$ we have $0 < f'(x)= \ln{\sqrt{2}}\left( \sqrt{2}\right)^x \leq 2 \ln{\sqrt{2}} < 2 \cdot \frac{2}{5}= \frac{4}{5}$
$$0 \leq 2 - x_{n+1} = \left( \sqrt{2}\right)^2 - \left( \sqrt{2}\right)^{x_n} = f'(\xi_n)(2-x_n) < \frac{4}{5}(2-x_n)$$
It follows
$$0 \leq 2 - x_{n+1} < \left(\frac{4}{5}\right)^n (2-x_1)\stackrel{n \to \infty}{\longrightarrow} 0 \Rightarrow  \boxed{\lim_{n \to \infty}x_n = 2}$$
To sum this up:


*

*$f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

*The fixpoint is unique as $|f'(x)| \leq q < 1$ on $[1,2]$.

*For any starting value $x_1 \in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]\Leftrightarrow x=2$
A: As the exponential is a growing function,
$$x<2\implies \sqrt2^{\,x}<\sqrt2^{\,2}=2$$ and the sequence starting from $x=\sqrt2<2$ is bounded above.
A: To find a candidate for the limit you have the equation
$$x=(\sqrt 2)^x\tag1$$
for some possible limit $x:=\lim_{n\to\infty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=\sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.
Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done. 
REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.
