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$$ f: (0,\infty)$$ is bounded and$$ f(x)>0 $$ How to prove that: If $$\lim_{x\to \infty} \frac{f(x+1)}{f(x)} = g$$, Then $$\lim_{x\to \infty} f(x)^{1/x} = g $$

I remember a similary proof for a strings using inequality between average

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Perhaps you can break it down into these: $$ \liminf \frac{f(x+1)}{f(x)} \le \liminf f(x)^{1/x} \le \limsup f(x)^{1/x} \le \limsup \frac{f(x+1)}{f(x)} $$


and then, if it doesn't work, analyze it to get a counterexample.

For $x>0$ write $x = n + t$ with $n \in \mathbb N$ and $0<t\le 1$. So $t$ is the fractional part (but not allowed to be zero). Define $$ f(x) := t\cdot 2^{-x}. $$

f $f(x)$

Compute $$ \frac{f(x+1)}{f(x)} = \frac{t\cdot 2^{-(x+1)}}{t\cdot 2^{-x}} = \frac{1}{2} $$ so certainly converges to $g:=1/2$. But I claim that $f(x)^{1/x}$ does not converge to $1/2$.

f^1/x $f(x)^{1/x}$

Indeed, given any $n \in \mathbb N$ there is $t_0\in (0,1)$ so that $t_0^{1/(n+1)} = 1/2$. Namely $t_0 = (1/2)^{n+1}$. But then for any $y$ there is $x = n+t_0 > y$ where $$ f(x)^{1/x} = \left(t_0 \cdot 2^{-x}\right)^{1/x} = t_0^{1/x}\cdot 2^{-1} < t_0^{1/(n+1)}\cdot \frac{1}{2} = \frac{1}{4} . $$

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This proof is incomplete...

$$\ln f(x)^{\frac{1}{x}}= \frac{\ln(f(x))}{x}$$

Then by Stolz Cezaro

$$\lim_x \ln f(x)^{\frac{1}{x}} = \lim_x \frac{\ln(f(x))}{x}=\lim_x \frac{\ln(f(x+1))-\ln(f(x))}{x+1-x}= \lim_x \ln (\frac{f(x+1)}{f(x)})$$

The issue is the fact that the equality

$$\lim_x \frac{\ln(f(x))}{x}=\lim_x \frac{\ln(f(x+1))-\ln(f(x))}{x+1-x}$$

holds for sequences, not for continuous $x$, but should be easy to prove the same way...

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"...should be easy." Maybe some sketch of proof would be nice, at least. – Pedro Tamaroff Mar 15 '13 at 21:18

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