How find this limit $\lim_{n\to\infty}a_{n}$ 
let $f(x)=x\ln{x}  (x>0)$,
  and $f_{1}(x)=f(x)$,and such $f_{2}(x)=f(f_{1}(x)),f_{3}(x)=f(f_{2}(x)),\cdots,f_{n+1}(x)=f(f_{n}(x))$

Assume that the sequnce $\{a_{n}\}$ such  $f_{n}(a_{n})=1$
Find the $$\lim_{n\to\infty}a_{n}$$
My try: since

$$f_{2}(x)=f(f_{1}(x))=f(x\ln{x})=x\ln{x}\ln{(x\ln{x})}=x\ln^2{x}+x\ln{x}\ln{(\ln{x})}$$
  $$f_{3}(x)=f(f_{2}(x))=f(x\ln^2{x}+x\ln{x}\ln{(\ln{x})})=[x\ln^2{x}+x\ln{x}\ln{(\ln{x})}]\ln{[x\ln^2{x}+x\ln{x}\ln{(\ln{x})}]}=\cdots\cdots$$
  and I can't work,But I fell guess
  $$\lim_{n\to\infty}a_{n}=e?$$

 A: $f(x)=x\ln(x)$, and $f'(x)=\ln(x)+1$, So $f$ is an increasing function from $[1,\infty)$ to $[0,\infty)$. Note that $f(e)=e$, and on $(1,\infty)$, $f'(x)>1$. (Hence $e$ is the only possible fixed point).
Let g defined on $[0,\infty)$ by $g(f(x))=x$. $g$ is an increasing function from $[0,\infty)$ to $[1,\infty)$. As $g=f^{-1}$, then $g(e)=e$ and on $(0,\infty)$, $0<g'(x)<1$. (Hence $e$ is the only possible fixed point).
And obviously $x<g(x)<e$ on $(0,e)$.
$a_n=g^n(1)$. So for all $n$, $e>a_{n+1}>a_n$, so $a_n$ converges to a limit $L$ that must satisfy $L=g(L)$, Hence $L=e$.
A: You consider $f: [1,\infty) \to [0,\infty), f(x)=x\ln x$ which is increasing and $f'(x)=1+\ln x>1$ for every $x >1$.
$f_n(a_n)=1, f_n(a_{n+1})=f^{-1}(1)>1$ so we have $a_{n+1}>a_n$. $f_n(a_n)=1<e=f_n(e)$ so $a_n <e$. Therefore $a_n$ is convergent and $a_n \to L \in (1,e]$. 
It is not hard to see that $f(a_{n+1})=a_n$ (because $f_n$ is injective).
Apply the mean value theorem on $[a_{n+1},e]$ to get $|a_{n}-e|=|f(a_{n+1})-f(e)| =|f'(x_n)| |a_{n+1}-e|$ with $x_n \in (a_{n+1},e)$. In particular $x_n>a_1 \geq 1+\varepsilon$ for some $\varepsilon>0$. Therefore $|a_n-e| \geq (1+\varepsilon)|a_{n+1}-e|$. 
This gives $\frac{|a_n-e|}{|a_{n+1}-e|} \geq 1+\varepsilon$. If $L \neq e$ we have no indetermination, so as $n \to \infty$ we obtain $1 \geq 1+\varepsilon$. Contradiction. Therefore $L=e$.
