Proving Cauchy Theorem This theorem says that a function $f$ is defined in the interval (a, +$\infty$) and it is bounded in every finite interval (a,b).
Then it holds that: (assume that the limit of the RHS exists)

*

*$$\lim_{x\to \infty} \frac{f(x)}{x} = \lim_{x\to \infty}[f(x+1) - f(x)];$$

*$$\lim_{x\to \infty} [f(x)]^{1/x} = \lim_{x\to \infty}\frac{f(x+1)}{f(x)}.$$
My question is how to prove the above equations. Any help will be appreciated.
 A: First point: Suppose the limit of $f(x+1)-f(x)$ exists (let’s call it $c$). Then for $\epsilon>0$ there is some $R$ s.t. if $x>R$ then $f(x+1)-f(x)$ deviates from $x$ by at most $\epsilon$. Then we can set
$$ f(x) = f(x)-f(x-1)+f(x-1)-f(x-2)+\ldots+f(l) $$
where $l\in(a,a+1]$, so $|f(l)|<C_1$. Also we get that the part of the sum with arguments that are smaller than $R$ is bounded by some $C_2=C_2(R)=C_2(\epsilon)$.
Thus we have
$ f(x) $ is a sum of something bounded by $C_1+C_2$ and a sum formed by $f(y)-f(y-1)$ for some $y>R+1$, so each of these summands is about $c+\delta(y)$ with $|\delta(y)|<\epsilon$. Thus this second part is within $Nc \pm N\epsilon$, where $N$ is the number of such telescope terms in the second part. Note that the number of terms in the first part is bounded by $R-a$, so we have $N(x,\epsilon)/x\to 1$ at $x\to\infty$.
Thus $f(x)/x$ is within $N/x(c\pm\epsilon) \pm (C_1+C_2)/x$ and goes with $x\to\infty$ towards $c$.
The second point follows from the first point by taking the logarithm of both sides and then confirming that $\log\circ f$ still satisfies the conditions to $f$, which is the case as long as $f(x)>0$ and it does not get close to $0$ unless it is in the limit.
Note that if the limit of $f(x+1)/f(x)$ exists this implies that if $f(x_n)\to 0$ at $x_n\to x_0$ for $x_0$ sufficiently large this must also hold true at $x_n+k\to x_0+k$ (and to some extent at $x_n-k\to x_0-k$, as long as $x_0-k$ is large enough). Thus if we take a sufficiently large $x$ and remove all points $p$ from $(x,x+1]$ where $f(p)<\epsilon$ along with all $p+k$ we can ensure that there cannot be a bounded sequence left with $f(x_n)\to 0$.
So we can at least proof it for all of these. But then by having $\epsilon$ go to $0$ we get the result for all $x$ so that $f(x)>0$.
If $f(x)=0$ for infinitely many arbitrarily large $x$ the RHS does inherently make not sense, if $f(x)<0$ for such cases the LHS does not make sense. So it is reasonable to assume that $f(x)>0$ for sufficiently large $x$.
