Q:

asymptote problem?

I was curious about this exercise, because I thought it could be a valuable tool to use the theorem dell'asintoto ... I do not think that is the way, does anyone have any idea? !
Let $ h $ is a function defined on $(a, + \infty)$ and limited all intervals $(a, b)$
$a <$ b. Prove that if
$$ \lim_ {x \to + \infty} [h (x +1)-h (x) ] = k, $$
then $$\lim_ {x \to + \infty} \frac {h (x)}{x} =k$$

A:

Let $\epsilon>0$.  There is an $N$ such that if $x>N$, then $|h(x+1)-h(x)-k|<\epsilon$.  Therefore, $$|h(y+n)-h(y)-kn|=\left|\sum_{i=1}^{n}(h(y+i)-h(y+i-1))-k\right|\leq n\epsilon$$ for $y>N+1$.
Suppose that $A<h(x)<B$ when $y\in [N+1,N+2]$.  If $y\in [N+1+n,N+2+n]$, we then have $A-n\epsilon<h(y)-nk<B+n\epsilon$.  Adding $nk-yk$ and dividing by $y$ (assuming that $y>0$), we have
$$\frac{A-n\epsilon +(n-y)k}{y} < \frac{h(y)}{y}-k < \frac{B+n\epsilon +(n-y)k}{y}.$$
Since $y-n\in [N+1,N+2]$ and $\frac{n}{y}<\frac{y-N}{y}<1+\frac{|N|}{y}$, we therefore have
$$\left|\frac{h(y)}{y}-k\right| < \frac{\max(|A|,|B|)+|N|+2}{y}+\epsilon \left(1+\frac{|N|}{y}\right).$$
For large values of $y$,  $\frac{\max(|A|,|B|)+|N|+2}y<\epsilon$ and $1+\frac{|N|}y<2$, so 
$$\left|\frac{h(y)}{y}-k\right|<3\epsilon.$$
Since we can do this for every $\epsilon>0$, we must have $\displaystyle \lim_{x\to \infty} h(x)/x=k$.

