Proving a specific statement involving limits related to a function with some given properties. 
Let $n\in\Bbb N$. Let a function $f(x)$ be bounded in every interval $(x_0, x_1)$. The domain of $f(x)$ is $x\in (x_0, +\infty)$. Prove that if the following limit exists and is either finite or infinite:
  $$
\lim_{x\to+\infty}{f(x+1)-f(x)\over x^n}=L \in \mathbb R\ \cup \{\pm \infty\}\tag1
$$
  Then:
  $$
\lim_{x\to+\infty}{f(x)\over x^{n+1}} = {1\over n + 1}\lim_{x\to+\infty}\frac{f(x+1)-f(x)}{x^n} \tag 2
$$

I've started with considering the following. Since $f(x)$ is bounded on every interval then there exist a sequence bounding $f(x)$. Namely:
$$
|f(x)| \le M_1,\ x\in(x_0, x_1)\\
|f(x)| \le M_2,\ x\in(x_0, x_2)\\
\cdots\\
|f(x)| \le M_k,\ x\in(x_0, x_k)
$$
where:
$$
x_1 < x_2 < \cdots < x_k
$$
We may now choose limsup and liminf of $M_k$ to be bounds for $f(x)$. So
$$
\liminf_{k\to\infty} M_k \le f(x) \le \limsup_{k\to\infty} M_k \tag3
$$
Then I tried assuming that the limit in $(1)$ is finite. Namely:
$$
\lim_{x\to+\infty}{f(x+1)-f(x)\over x^n} = A \in \mathbb R
$$
Moreover, if the reasoning to arrive at $(3)$ is correct, then dividing $(3)$ by $x^{n+1}$ we obtain:
$$
\frac{\liminf_{k\to\infty} M_k}{x^{n+1}} \le \frac{f(x)}{x^{n+1}} \le \frac{\limsup_{k\to\infty} M_k}{x^{n+1}}
$$
LHS and RHS of the inequality are finite numbers so taking the limit in that inequality I got:
$$
\lim_{x\to+\infty}\frac{\liminf_{k\to\infty} M_k}{x^{n+1}} \le \lim_{x\to+\infty}\frac{f(x)}{x^{n+1}} \le \lim_{x\to+\infty} \frac{\limsup_{k\to\infty} M_k}{x^{n+1}}
$$
Which implies that:
$$
\lim_{x\to+\infty}\frac{f(x)}{x^{n+1}} = 0
$$
But that doesn't seem right. I'm lost here since I see no way to arrive at $(2)$. How can I prove the statement from the problem section?
 A: Is suffices to show that
$$
 \limsup_{x\to\infty}{\frac{f(x)}{x^{n+1}}} \le \frac 1{n+1} \limsup_{x\to\infty}\frac{f(x+1)-f(x)}{x^n} \tag{*}
$$
because then the same reasoning can be applied to $-f$, so that
$$
\frac 1{n+1}  \liminf_{x\to\infty}\frac{f(x+1)-f(x)}{x^n} \le \liminf_{x\to\infty}{\frac{f(x)}{x^{n+1}}} \le \limsup_{x\to\infty}{\frac{f(x)}{x^{n+1}}} \le \frac 1{n+1}  \limsup_{x\to\infty}\frac{f(x+1)-f(x)}{x^n}
$$
and the desired conclusion follows.
In order to prove $(*)$ we can assume that
$$
 L = \limsup_{x\to\infty}\frac{f(x+1)-f(x)}{x^n} < \infty \, .
$$
For every $K > L$ there is a $c > x_0$ such that 
$$
 f(x+1) - f(x) \le K x^n \text{ for } x \ge c \, .
$$
Let $k = \lfloor x-c \rfloor$. Then $c \le x-k \le c+1$ and
$$
 f(x) \le f(x-k) + \sum_{j=1}^k f(x-j+1) - f(x-j) 
\le f(x-k) + K \sum_{j=1}^k (x-j)^n \\
\le f(x-k) + K \int_{x-k}^{x} t^n \, dt = f(x-k) + \frac{K}{n+1} (x^{n+1} - (x-k)^{n+1}) \\
\le  f(x-k) + \frac{Kx^{n+1}}{n+1} \\
\le \max \{ f(t) : c \le t \le c+1 \} + \frac{Kx^{n+1}}{n+1} \,.
$$
It follows that
$$
\limsup_{x\to\infty}{\frac{f(x)}{x^{n+1}}} \le \frac{K}{n+1} \, .
$$
This holds for every $K > L$, therefore
$$
\limsup_{x\to\infty}{\frac{f(x)}{x^{n+1}}} \le \frac{L}{n+1} 
$$
and the proof of $(*)$ is finished.
Remark #1: The above reasoning is similar to and inspired by the proof of the Stolz–Cesàro theorem. I was not able to apply that theorem directly.
Remark #2: The problem with your approach is that $\limsup_{k\to\infty} M_k$ need not be finite.
