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

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 $$(x-\delta_\epsilon)(1-\epsilon)\delta_\varepsilon.$$
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.

• There should be another condition also that $f$ is bounded on every bounded interval. Commented Sep 20, 2021 at 10:39
• The problem is that your argument works only when $x=\delta_{\epsilon} +n$ where $n$ is a positive integer. For it to work for all $x>\delta_{\epsilon}$ you will need $f$ to be bounded in $[\delta_{\epsilon}, \delta_{\epsilon} +1]$. I hope you understand the flaw. Commented Sep 20, 2021 at 11:10
• @ParamanandSingh Ah, thanks, it does make sense, I felt that the way I tried to mimick what I would do for a sequence may be flawed. Now I understand, thank you very much! Commented Sep 20, 2021 at 11:14
• @ArcticChar: things happen so fast here and I am limited in my speed due to mobile device. :) Commented Sep 20, 2021 at 11:25
• @ParamanandSingh: 1) For me as a reader of SE, it would be quite confusing: Following a [duplicate] link that deviates in non-minor details. The answer to that other question cannot be applied to this one. 2) It's still preferable to have a proper answer (instead of having to dig into every comment quere). Afterall, the two proofs (one confirming, one counter-example) are quite different. 3) It's nice that the TO found help and can proceed; but IIUC, SE is also about sharing information with "passive" readers. Otherwise, all except the questions could be private. Commented Sep 20, 2021 at 11:32

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.

• thanks, so I guess that my proof fails here in this case: $\displaystyle \limsup_{x\to \infty}\frac{f(\delta_\epsilon)+(x-\delta_\epsilon)(1-\epsilon)}{x}$ (I don't really see any other place where it might). I thought that as $x\to \infty$ $f(\delta_\epsilon)$ would stay constant. Commented Sep 20, 2021 at 11:01
• +1 for the fine counterexample. Commented Sep 20, 2021 at 11:29
• @TheZone I think that's where your error is, yeah.
– 5xum
Commented Sep 20, 2021 at 11:34

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.

• thanks, now it is all really clear. Commented Sep 20, 2021 at 12:14
• :P Wasn't sure it was already clear as you just assumed "I don't really see any other place where it might". Commented Sep 20, 2021 at 12:23

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 $$\lim_{x\to 0}\frac{f(x)}{x}=0.$$

• Your third equality should be $-\sin(\pi \lfloor x \rfloor)=0$ not $1$. Btw: $\sin(x)\cos(x)=\sin(2x)/2$ so actually $f(x)=0$ for all $x$. Commented Sep 20, 2021 at 15:48