The following problem is from Chapter 22 "Infinite Sequences", from Spivak's Calculus

17.(a) Prove that if $\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$, then $\lim\limits_{n\to\infty} \frac{a_n}{n}=l$.

(b) Suppose that $f$ is continuous and $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$. Prove that

$$\lim\limits_{x\to\infty} \frac{f(x)}{x}=l$$

Hint: Let $a_n$ and $b_n$ be the $\inf$ and $\sup$ of $f$ on $[n,n+1]$.

I am interested in item $(b)$.

The solution manual has a solution that uses a different path than I try below (though I don't ultimately succeed). I have not analyzed the solution manual solution in depth yet, though I know it uses the given hint.

My question is if the path in the solution below could be finished in some way.

Here was my initial (incomplete) attempt at a solution

Previously (Chapter 13, Problem 40), we proved

$f$ integrable on $[0,x]$ for all $x>0$ and $\lim\limits_{x\to\infty} f(x)=a$, then

$$\lim\limits_{x\to\infty} \frac{1}{x}\int_0^x f=a$$

Let's use this result.

Let $g(x)=f(x+1)-f(x)$.

$g$ is

  • continuous and hence integrable on $[0,x]$ for all $x>0$.
  • $\lim\limits_{x\to\infty} g(x)=l$

By 13-40,

$$\lim\limits_{x\to\infty} \frac{1}{x}\int_0^x g=l$$


$$\int_0^x g=\int_0^x f(t+1)dt-\int_0^x f(t)dt$$

$$=\int_1^{x+1} f-\int_0^x f=\int_x^{x+1}f-\int_0^1 f$$


$$\lim\limits_{x\to\infty} \frac{1}{x}\int_0^x g = \lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1}f-\lim\limits_{x\to\infty}\frac{1}{x}\int_0^1 f$$

$$=\lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1} f=l$$

By the Mean Value Theorem for Integrals, for each $x$ there is some $t\in [x,x+1]$ such that

$$f(t)=\int_x^{x+1} f$$


$$\lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1} f=\lim\limits_{x\to\infty}\frac{f(t(x))}{x}=l\tag{1}$$

The question is how to get from $(1)$ to


One attempt starts from the definition of limit

$$\left | \frac{f(t(x))}{x}-l \right |<\epsilon$$

If I were able to say that $|f(t(x))-f(x)|$ is bounded for all $x$ I could maybe use the argument that, if $m$ and $M$ are such lower and upper bounds

$$\frac{f(x)}{x}+\frac{m}{x}\leq \frac{f(t(x))}{x}=\frac{f(x)}{x}+\frac{f(t(x))-f(x)}{x}\leq\frac{f(x)}{x}+\frac{M}{x}$$

$$\lim\limits_{x\to\infty}\left [\frac{f(x)}{x}+\frac{m}{x}\right ] \leq \lim\limits_{x\to\infty}\frac{f(t(x))}{x}\leq\left [\lim\limits_{x\to\infty}\frac{f(x)}{x}+\frac{M}{x}\right ]$$

$$\lim\limits_{x\to\infty}\frac{f(x)}{x} \leq l\leq\lim\limits_{x\to\infty}\frac{f(x)}{x}$$

However, I'm not sure I can use this argument. I would need to prove that $|f(t)-f(x)|$ is bounded for all $x$.

Intuitively, it seems that in the limit, the function has the same shape between any two points. But this means that it is bounded between any two points, since it is continuous.

  • $\begingroup$ Why $\int_{1}^{x+1}-\int_{0}^{1}$ would become $\int_{x}^{x+1}-\int_{0}^{1}$? $\endgroup$
    – user284331
    Oct 20, 2022 at 0:06
  • $\begingroup$ @user284331 there was a typo. It should be $\int_1^{x+1}-\int_0^x$ that becomes $\int_x^{x+1}-\int_0^1$. $\endgroup$
    – evianpring
    Oct 20, 2022 at 0:11
  • $\begingroup$ See math.stackexchange.com/q/192963/72031 $\endgroup$
    – Paramanand Singh
    Oct 20, 2022 at 1:49

2 Answers 2


There is some room for interpretation in answering your question.

Let me start by showing the following. Assume $f$ is continuous and it satisfies $\lim_{x\to\infty} \frac{1}{x}\int_x^{x+1} f(t)dt=\ell$. This is NOT sufficient to argue that $\lim_{x\to\infty} \frac{f(x)}{x}=\ell$.

Indeed, consider the function $f(x)=x\sin{2\pi x}$. It is not difficult to see by integrating by parts that $\int f(x)dx= \frac{1}{4\pi^2}(\sin{2\pi x}-2\pi x\cos{2\pi x})+C$, so $\int_x^{x+1} f(t)dt=\frac{1}{2\pi} \cos{2\pi x}$. Therefore $\ell=0$ in this case. However, $\lim_{x\to\infty} \frac{f(x)}{x}=\lim_{x\to\infty} \sin{2\pi x}$ does not exist.

So if your approach only uses the assumption that $\lim_{x\to\infty} |f(x+1)-f(x)|=\ell$ to get the limit for the integral, it will not work. Note that with my example, this limit is again exactly the limit of $\sin{2\pi x}$ which does not exist.

Therefore, if you want to take this approach, the question boils down to being able to use the existence of the limit of $|f(x+1)-f(x)|$ again after the integral limit in a way that is different from the suggested textbook proof. If you cannot do that, then probably this is a dead end.


As you said, it is not clear if the function $f(t(x))-f(x)$ is bounded on $\mathbb{R}$.

Anyway, if you assume further that $f'$ is continuous, then I have some sort of solution similar to your idea:

Write \begin{align*} f(x+1)-f(x)=\int_{x}^{x+1}f'(t)dt. \end{align*} Then \begin{align*} \frac{1}{x}\int_{0}^{x}\int_{z}^{z+1}f'(t)dtdz\rightarrow l. \end{align*} But \begin{align*} \frac{1}{x}\int_{0}^{x}\int_{z}^{z+1}f'(t)dtdz&=\frac{1}{x}\int_{0}^{1}\int_{0}^{1}f'(t)dzdt+\frac{1}{x}\int_{1}^{x}\int_{t-1}^{t}f'(t)dzdt\\ &=\frac{1}{x}\int_{0}^{1}\int_{0}^{1}f'(t)dzdt+\frac{1}{x}\int_{1}^{x}\int_{t-1}^{t}f'(t)dzdt\\ &=\frac{1}{x}\int_{0}^{1}\int_{0}^{1}f'(t)dzdt+\frac{1}{x}\int_{1}^{x}f'(t)dt\\ &=\frac{1}{x}\int_{0}^{1}\int_{0}^{1}f'(t)dzdt+\frac{1}{x}(f(x)-f(1)), \end{align*} then it is clear that $f(x)/x\rightarrow l$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .