How to prove that $\lim\limits_{x\to\infty}f(x)^{1/x}=g$ if $\lim\limits_{x\to\infty}\frac{f(x+1)}{f(x)}=g$ $$ f: (0,\infty)$$ 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
 A: 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
$$
f(x) := t\cdot 2^{-x}.
$$
 $f(x)$
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 $1/2$.  
 $f(x)^{1/x}$
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
$$
f(x)^{1/x} = \left(t_0 \cdot 2^{-x}\right)^{1/x} = t_0^{1/x}\cdot 2^{-1}
< t_0^{1/(n+1)}\cdot \frac{1}{2} = \frac{1}{4} .
$$
A: 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...
