$$ f: (0,\infty)$$ 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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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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Perhaps you can break it down into these: $$ \liminf \frac{f(x+1)}{f(x)} \le \liminf f(x)^{1/x} \le \limsup f(x)^{1/x} \le \limsup \frac{f(x+1)}{f(x)} $$ ...added... 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 $$ f(x) := t\cdot 2^{-x}. $$
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 $1/2$.
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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$f(x)$
$f(x)^{1/x}$
