It goes on and on

Calculus Level 3

$\large L = \lim_{x\to0^+} x^{x^{x^{x^{\cdot^{\cdot^\cdot}}}}}$

Find the value $L$ when the number $x$ 's in the infinitely nested function above is even.

0 1

\pi

\infty

1 solution

Bar Hemo
Apr 18, 2016

Very easy to prove with induction.

base: n=2

$\lim_{x \to 0+} x^{x}=\lim_{x \to 0+} e^{xln(x)}=l'hopital \: rule=1$

assume it works for n=2k

$\lim_{x \to 0+} x^{x^{x^{x^{...}}}}= 2k \: x's= 1= f(x)$

step: we will prove for n=2(k+1)

$\lim_{x \to 0+} x^{x^{x^{x^{...}}}}=\lim_{x \to 0+} x^{x^{f(x)}}=\lim_{x \to 0+} x^{e^{f(x)ln(x)}}=1$

The algebra in your induction step is flawed since $x^{x^{x^x}}\neq (x^x)^{x^x}$

Otto Bretscher - 5 years, 1 month ago

I think it is fixed now.

Bar Hemo - 5 years, 1 month ago

I'm still not fully convinced. I agree that $\lim_{x \to 0^+}e^{f(x)\ln(x)}=0$ ... but then we have another indeterminate form $0^0$ on our hands.

There is another, deeper issue with your solution: You are attempting to show that $\lim_{x\to 0^+}x^{x^{.^{.^x}}}=1$ for any finite power tower with a given even number of $x$ 's. But you are claiming much more: You are implicitely claiming that the infinite power tower converges (for all $x$ ?) and that the limit of the value of that infinite power tower is 1 as $x\to 0^+$ .

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – I ment that you can place 1 instead of f(x) and than its $x^x$ again. But Im not sure about it.

Bar Hemo - 5 years, 1 month ago

@Bar Hemo – I'm afraid we are not allowed to do that ;)

Otto Bretscher - 5 years, 1 month ago

It goes on and on

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