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I would like to know if it is possible generalize this result and how it could be shown:

we know that

$$ \begin{align} \lim_{x\to 0^+}x^x & =1;\\ \\ \lim_{x\to 0^+}x^{x^x} & =0;\\ \\ \lim_{x\to 0^+}x^{x^{x^x}}& =1 \end{align} $$

$$\begin{matrix}\displaystyle\lim_{x\to 0^+} \overbrace{x^{x^{x^{x^{\cdots^{x}}}}}}^{n\text{ times}} \end{matrix}\quad$$

I would like to conclude that if $n$ equal limit is $1,$ and if $n$ is odd limit is $0$ is it Possible?

tanks in advances

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3  
Yes. Use induction. –  Ragib Zaman Mar 17 '12 at 15:11
    
yes , but in wich way? –  FrConnection Mar 17 '12 at 15:20
    
@RagibZaman half of the induction is not trivial, since $\lim_{(x,y)\rightarrow(0+,0+)} x^y$ does not exist –  Thomas Andrews Mar 17 '12 at 15:24

1 Answer 1

up vote 1 down vote accepted

We can proove it by induction on $n$ :

Let us consider the sequence of function $0 \leq a_n(x) = x^{a_{n-1}(x)} \leq 1$.

Initialisation is OK for $n=0$.

If $n$ is odd then $a_n(x) = x^{a_{n-1}(x)}$ by induction hyposthesis $a_{n-1}(x) \rightarrow 1$ then for all $ 1/2\geq a \geq 0$ there exist $x_0$ s.t. if $x \geq x_0$ we have $a_n(x) \leq x^{1-a} \rightarrow 0$ because the fonction $ x \rightarrow c^x$ is decreasing on $(0,1)$ with $c \in (0,1)$.

If $n$ is even $a_n(x) = x^{x^{a_{n-2}(x)}} \geq x^{x^{1-a}} $ for the same reason above, for $0\leq a \leq 1/2$. But $x^{x^{1-a}} = e^{x^{1-a}ln(x)} \rightarrow 1 $

And that complete the induction and the proof.

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