If $ \lim_{x\to \infty} (f(x+1)-f(x))=1$, then $ \lim_{x\to \infty} \frac{f(x)}{x}=1$?

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a fuction such that $\displaystyle \lim_{x\to \infty} (f(x+1)-f(x))=1$. Is it true then that $\displaystyle \lim_{x\to \infty} \frac{f(x)}{x}=1$?

I think it is and here is how I went about it. Let $\varepsilon>0$. Then there is some $\delta_\varepsilon>0$ such that $$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon, \forall x>\delta_\varepsilon.$$ Thus, we may write the following inequalities for an $x> \delta_\epsilon$:
$$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon\\ 1-\varepsilon < f(x-1)-f(x-2)<1+\varepsilon\\ \vdots\\ 1-\varepsilon < f(\delta_\varepsilon+1)-f(\delta_\varepsilon)<1+\varepsilon$$
and after we sum these up we get that $$(x-\delta_\epsilon)(1-\epsilon)<f(x)-f(\delta_\varepsilon)<(x-\delta_\varepsilon)(1+\varepsilon), \forall x>\delta_\varepsilon.$$
This implies that $$\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1-\varepsilon)}{x}<\frac{f(x)}{x}<\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1+\varepsilon)}{x}, \forall x>\delta_\varepsilon.$$
If we take $\displaystyle\limsup_{x\to\infty}$, we get that $1-\varepsilon < \displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}< 1+\varepsilon$, $\forall \varepsilon > 0$, so $\displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}=1$. In the same way we get that $\displaystyle\liminf_{x\to\infty} \frac{f(x)}{x}=1$, so $\displaystyle\lim_{x\to\infty} \frac{f(x)}{x}=1$ as desired.

Is this proof correct? I am a bit unsure that I am taking that $\limsup$ correctly, even though I can't see why it could be wrong.

Answer 1

Take the function $$f(x)=\begin{cases}\frac{1}{[x]} + \lfloor x\rfloor;& x\notin \mathbb Z\\ \lfloor x\rfloor ; &x\in\mathbb Z\end{cases}$$

where $[x]$ denotes the fractional part of $x$, i.e. $[x]=x-\lfloor x\rfloor $.

Then the function satisfies your condition, since $f(x+1)-f(x)=1$ is true for all $x$. However, the limit $$\lim_{x\to\infty}\frac{f(x)}{x}$$ does not exist, because the value of $f$ on every interval $[n, n+1]$ is unbounded.

Answer 2

In your proof you are implicitly assuming that $x$ is such that $x-\delta_\epsilon=n$ is an integer $n$, so you have some constraint on the specific value of $x$ as $x\rightarrow \infty$. In the example given by 5xum this means that the term $\frac{1}{[x]}=\frac{1}{[\delta_\epsilon]}$ which you assume to be fixed. If you want to allow general $x$ you would have to modify your inequality to $$\frac{f(x-\lfloor x-\delta_\epsilon\rfloor)+\lfloor x-\delta_\epsilon\rfloor(1-\epsilon)}{x}<\frac{f(x)}{x}<\frac{f(x-\lfloor x-\delta_\epsilon\rfloor)+\lfloor x-\delta_\epsilon\rfloor(1+\epsilon)}{x}, \forall x>\delta_\epsilon.$$ Now you have to assume that $f(x-\lfloor x-\delta_\epsilon \rfloor)$ is bounded $\forall x>\delta_\epsilon$ in which case your proof would be correct. This however is not the case in 5xum's example.

Answer 3

Let $ f(x)= -\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right). $ Then \begin{align} f(x+1)-f(x) =& \sin \left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right) - \sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \\ =&+\sin\left((\lfloor x+1\rfloor - \lfloor x\rfloor)\cdot \dfrac{\pi}{2}\right) \\ =& +\sin\left( 1\cdot \frac{\pi}{2} \right) \\ =& 1 \end{align} and by $-1\leq f(x)=- \sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right) \cdot \cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)\leq +1$ we have $$ \lim_{x\to 0}\frac{f(x)}{x}=0. $$