Evaluating $\lim\limits_{x\rightarrow \infty} \frac{f(x)}{x}$ knowing $\lim\limits_{x\to\infty} f(x+1) - f(x)$ [duplicate]

Let $f : [0, +\infty)\to \mathbb{R}$ be a continuous function such that $$\lim_{x\to+\infty}f(x+1) - f(x) = a.$$ Prove that: $$\lim\limits_{x\rightarrow \infty} \dfrac{f(x)}{x} = a.$$

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marked as duplicate by Did, hardmath, user127.0.0.1, Git Gud, TMMJan 26 '14 at 20:21

for big 'x' $f(x+1)-f(x) \sim f'(x)$ – Jose Garcia Jan 26 '14 at 18:23
@JoseGarcia It wasn't assumed $f$ is differentiable. – David Mitra Jan 26 '14 at 18:24
In the case of sequences, Stolz-Cesaro. – Martín-Blas Pérez Pinilla Jan 26 '14 at 18:25
Everything works here: the direct epsilon-delta approach, or Cesaro, or any question on the site this one is a duplicate of. – Did Jan 26 '14 at 18:27
There is a solution to this problem here. math.stackexchange.com/questions/192963/… – TheNumber23 Jan 26 '14 at 18:27

HINT: $$\lim_{x\rightarrow +\infty}\frac{f(x)}x=\lim_{x\rightarrow +\infty}\frac{[f(x)-f(x-1)]+[f(x-1)-f(x-2)]+...}x$$

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Hint: Reduce to the case where $a=0$ by letting $g(x)=f(x)-ax$.

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In LeeNeverGup's proposition, and when $x$ goes to $+\infty$, you have, by summing the limits : $$\frac {f(x)}{x} \sim x*\frac ax$$

Then you can easily deduce that $\lim_{x\to 0} \frac{f(x)}{x} = a$ because $x\to +\infty$ and then is $\neq 0$

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