If $\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell,$ then $\lim\limits_{x\to+\infty}\frac{f(x)}x=\ell$.

Let $f:[0,+\infty)\to\Bbb R$ be a function bounded on each finite interval.

I want to show that if $\lim\limits_{x\to+\infty}[f(x+1)-f(x)]= L,$ then also $\lim\limits_{x\to+\infty}\dfrac{f(x)}x = L$

Solution 1:

Since $\lim\limits_{x\rightarrow+\infty}[f(x+1)-f(x)]= L$, for any $\epsilon$ we have some $y>0$ such that $$x> y\implies |f(x+1)-f(x)-L|<\epsilon$$ so for any $x>y$ we have $$\begin{align} \frac{f(x)}{x} &=\frac{f(x)-f\left(x-\lfloor x-y\rfloor\right)}{x}+\frac{f\left(x-\lfloor x-y\rfloor\right)}{x}\\ &=\frac{1}{x}\left(\sum\limits_{i=1}^{\lfloor x-y\rfloor}(f(x-i+1)-f\left(x-i\right))+f\left(x-\lfloor x-y\rfloor\right)\right)\\ &\leq\frac{1}{x}\left(\sum\limits_{i=1}^{\lfloor x-y\rfloor}(L+\epsilon)+\sup\limits_{z\in [0,y+1)}|f(z)|\right)\\ &\leq\frac{\lfloor x-y\rfloor}{x}(L+\epsilon)+\frac{1}{x}\sup\limits_{z\in [0,y+1)}|f(z)|\\ &\leq L+\epsilon+\frac{1}{x}\sup\limits_{z\in [0,y+1)}|f(z)|\\ \end{align}$$ which as $x\to+\infty$ approaches $L+\epsilon$. Letting $\epsilon\to 0$ gives the us $\limsup\limits_{x\to+\infty} \frac{f(x)}{x}\leq L$. A similar technique shows that $\liminf\limits_{x\to+\infty} \frac{f(x)}{x}\geq L$, finishing the proof.