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Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b>1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.

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    Provided both exist, of course? Are you sure there aren't any additional hypotheses? – 2017-01-29
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    I'm looking into the book right now, and typed the problem as it is. – 2017-01-29
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    It would be enough to prove that $f(x)-L\,x$ has a limit at $\infty$ (which amounts to $f(x)$ having an oblique asymptote) but it's not entirely obvious that that's the case with just the given conditions. +1 for the question. – 2017-01-29
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    Thank you, I find this question nice too. – 2017-01-29

2 Answers 2

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The right statement is as follows:

  • If $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)=\ell\in\mathbb{R}$ then $\lim\limits_{x\to\infty}\dfrac{f(x)}{x}=\ell$.

Indeed, suppose that $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)=\ell\in\mathbb{R}$. Let $\varepsilon>0$, then there exists $b_\varepsilon>1$ such that $$\forall\,t\ge b_\varepsilon,\qquad -\varepsilon0$ is arbitrary. Therefore $$\ell \le \liminf_{x\to\infty}\frac{f(x)}{x}\le \limsup_{x\to\infty}\frac{f(x)}{x} \le\ell$$ That is the limit $\lim\limits_{x\to\infty}\dfrac{f(x)}{x} $ exists and is equal to $\ell$.

  • Note that the example $f(x)=\sin(\pi x)$ shows that $\lim\limits_{x\to\infty}\dfrac{f(x)}{x}$ might exist while $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)$ does not.
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    Thank you for clear and complete proof and for the counterexample. Even I was able to understand everything. – 2017-01-30
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    @AndreiPetrov, You are welcome.. – 2017-01-30
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If both exists :

$f$ bounded imply that $\lim_{x\to + \infty} \frac{f(x)}{x} = 0$

Now, suppose that $f(x)-f(x+1)$ doesn't converge to $0$

We have that

$$f(n) - f(0) = \sum_{k=0}^{n-1} f(k+1)-f(k)$$

As $ \lim_{k\to \infty} f(k+1)-f(k) $ exists (without loss of generality suppose it greater than $0$) and is not equal to $0$, there exists $\eta >0$ and $M>0$ such that $\forall n > M, f(n+1)-f(n) \geq \eta$

Hence

$$f(n) = \underbrace{f(0) + \sum_{k=0}^M f(k+1)-f(k)}_{Constant} + \sum_{k=M+1}^{n-1} f(k+1)-f(k)$$

$$f(n) \geq C_{\eta} + \sum_{k=M+1}^{n-1} \eta = C_\eta + (n-M+1) \eta$$

The left side is bounded and the right side diverge to $+\infty$ : contradiction

Now, it's easy to construct functions $f$ where $\lim_{x\to + \infty} \frac{f(x)}{x} $ exists and not $\lim_{x\to + \infty} f(x+1)-f(x) $

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    $f$ is only bounded on finite intervals $(1,b)\,$, not necessarily on the entire $(1,\infty)\,$. For example, $f(x)=x$ satisfies both the premise and conclusion of the problem. – 2017-01-29
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    Thank you for the solution, but this is only part of the problem. As per dxiv, $f(x)$ does not have to be bounded on the whole interval $(1,\infty)$, so, taking $f(x)=mx, m\in R$, we can have not only zero, but any real number as the limit. – 2017-01-29
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    @dixv : you're right, I didn't pay sufficent attention to the exact question. So what I wrote is kinda irrelevant to the current question. – 2017-01-29
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    @Tryss The idea may be salvageable if applied to $g(x) = f(x) - L\,x$ but I haven't thought it through. – 2017-01-30