$\lim x_n^{x_n}=4$ prove that $\lim x_n=2$ Let $(x_n)$ be a sequence of real numbers, such that:
$\lim x_n^{x_n}=4$, prove that $\lim x_n=2$
I'm not sure if my proof is right.
I assumed that $\lim x_n $ isn't 2 and using Cauchy's criterion:
$|x_n-2|>\epsilon$  so $ x_n>\epsilon+2$  or $x_n<-\epsilon+2$
$|x_n^{x_n}-4|<\epsilon $ so $x_n<\sqrt[x_n]{\epsilon+4}$
and then we combine what we've found and get:
$\epsilon+2<\sqrt[x_n]{\epsilon+4}$ 
$\epsilon+4<(\epsilon+2)^2<(\epsilon+2)^{\epsilon+2}<(\epsilon+2)^{x_n}<\epsilon+4$  and it's not true so  $\lim x_n=2$.
Is that okay?
Edit: I just wanted to know if my solution was right but the other post helped as well, thanks.
 A: There are interesting elements in your proof. However, the way you state it can be improved:


*

*You mention Cauchy criteria, but what you use is not Cauchy criteria.

*You suppose that $\lim x_n$ is not equal to $2$. You cannot say $|x_n-2|>\epsilon$ without mentionning of what $n$ your're speaking. You should state that you have a subsequence $(x_{\beta_n})$ for which $|x_{\beta_n}-2|>\epsilon$.

*From there you have a subsubsequence with either $x_{\beta_n^\prime}>\epsilon+2$  or $x_{\beta_n^\prime}<-\epsilon+2$.

*Then you can follow with your other arguments.


However, I think that a totally different proof can use the map $f : x \mapsto x^x$, that is continuous, strictly increasing for $x \in (1,+\infty)$ and therefore invertible around $x=2$ with a continuous inverse. The conclusion is then straightforward.
A: The function $g:\left(0,\infty\right)\rightarrow\mathbb{R}$ prescribed
by $x\mapsto x^{x}$ is continuous.
If $\left(x_{n}\right)$ is a sequence in $\left(0,\infty\right)$
with $\lim_{n\rightarrow\infty}g\left(x_{n}\right)=4$ then eventually
$1\leq x_{n}\leq3$ so that the sequence has convergent subsequences
that have limits in $\left[1,3\right]$.
If $\left(x_{n_{k}}\right)$ is an arbitrary convergent subsequence having $x\in\left[1,3\right]$
as limit then the continuity of $g$ assures that $g\left(x\right)=4$
hence $x=2$.
Now the conclusion is allowed that $\left(x_{n}\right)$ itself is
convergent and has $2$ as limit.
