For a non-negative integer $k$, $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$ Suppose that $f$ defined on $(a,\infty)$ is bounded on each finite interval $(a,b),a>b$. For a non-negative integer $k$, it is to be shown that $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$
Given an $\displaystyle \epsilon  >0,$there exists an integer $\displaystyle N$ such that
$$\displaystyle 
x \geq N \Longrightarrow \ \left| \frac{f( x+1) -f( x)}{x^{k}}
 -l \right| < \epsilon $$
Noting that
\begin{align*}
f( x) & =\sum _{i=1}^{[ x]} f( x-[ x] +i) -f( x-[ x] +i-1) +\color{red}{f( x-[ x])}
\end{align*}
it follows that
\begin{align*}
f( x) & =\sum _{i=1}^{N}\overbrace{( f( x-[ x] +i) -f( x-[ x] +i-1))}^{h( i)} +\sum _{i=N+1}^{[ x]}( f( x-[ x] +i) -f( x-[ x] +i-1)) +\color{red}{( \ )}
\end{align*}
It follows that
\begin{align*}
|\frac{f( x)}{x^{k+1}} -\frac{l}{k+1} | & \leq \sum _{i=1}^{N} |\frac{h( i)}{x^{k+1}} -\frac{l}{[ x]( k+1)} |+\sum _{i=N+1}^{[ x]} |\frac{h( i)}{x^{k+1}} -\frac{l}{[x](k+1)} |+\color{red}{( )x^{-k-1}}\\
 & \leq {\textstyle \frac{\overbrace{M}^{\sup _{1\leq i\leq N} h( i)}}{x^{k+1}} +\frac{N|l|}{[ x]( k+1)} +\frac{\epsilon ([ x] -N)}{x} +|\frac{l}{x} -\frac{l}{[ x]( k+1)} |([ x] -N) +\color{red}{\overbrace{\color{red}{\sup _{t\in [ 0,1)}f(t)}}^{M'} x^{-k-1}}}\\
 & =\frac{M+M'}{x^{k+1}} +{\textstyle \frac{N|l|}{[ x]( k+1)} +\color{purple}{|\frac{l[ x]}{x} -\frac{l}{( k+1)} |} -N|\frac{l}{x} -\frac{l}{[ x]( k+1)} |+\frac{\epsilon ([ x] -N)}{x}} \tag 1
\end{align*}
In $(1)$, except the purple term every other term can be made arbitrarily small. How do I take care of the purple term so that I can conlude the desired result by limit definition.

First we prove the statement for the case $l=0$, i.e.
$
\lim_{n \to \infty}\frac{f(x+1)-f(x)}{x^k} = 0
$.
Given $\epsilon > 0$ there is a $y > a$ such that for all $x \ge y$:
$$
\left | \frac{f(x+1)-f(x)}{x^k} \right | < \epsilon \, .
$$
Similarly as in If $\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell,$ then $\lim\limits_{x\to+\infty}\frac{f(x)}x=\ell$. we write
$$
 \frac{f(x)}{x^{k+1}} = 
\frac{1}{x}\left(\sum_{i=1}^{\lfloor x-y\rfloor}\frac{f(x-i+1)-f(x-i)}{(x-i)^k} \cdot \left( 1-\frac ix\right)^k
\right)+\frac{f(x-\lfloor x-y\rfloor)}{x^{k+1}}
$$
which implies
$$
\left| \frac{f(x)}{x^{k+1}}\right| \le \frac{\lfloor x-y\rfloor}{x} \epsilon + \frac{M}{x^{k+1}}
$$
with $M = \sup \{ |f(z)| : y \le z < y+1\}$. It follows that
$$
 \limsup_{n \to \infty }\left| \frac{f(x)}{x^{k+1}}\right| \le \epsilon \, .
$$
This holds for all $\epsilon > 0$, and therefore
$$
\lim_{n \to \infty} \frac{f(x)}{x^{k+1}} = 0 \, .
$$

For the general case with $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l$ we set
$$
 g(x) = f(x) - \frac{x^{k+1}}{k+1} l \, .
$$
$g$ is bounded on every finite interval $(a, b)$ and satisfies
$$
 \frac{g(x+1)-g(x)}{x^k} = \frac{f(x+1)-f(x)}{x^k} - l \frac{(x+1)^{k+1}-x^{k+1}}{(k+1)x^k} \\
 = \frac{f(x+1)-f(x)}{x^k} - l \frac{(1+1/x)^{k+1}-1}{(k+1)/x}
\to l - l = 0 
$$
and then
$$
 \frac{f(x)}{x^{k+1}} = \frac{g(x)}{x^{k+1}} + \frac{l}{k+1} \to 0 + \frac{l}{k+1} = \frac{l}{k+1} \, .
$$
