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,
$$\lim\limits_{x\to\infty} \frac{1}{x}\int_0^x g=l$$
Now,
$$\int_0^x g=\int_0^x f(t+1)dt-\int_0^x f(t)dt$$
$$=\int_1^{x+1} f-\int_0^x f=\int_x^{x+1}f-\int_0^1 f$$
Thus
$$\lim\limits_{x\to\infty} \frac{1}{x}\int_0^x g = \lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1}f-\lim\limits_{x\to\infty}\frac{1}{x}\int_0^1 f$$
$$=\lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1} f=l$$
By the Mean Value Theorem for Integrals, for each $x$ there is some $t\in [x,x+1]$ such that
$$f(t)=\int_x^{x+1} f$$
Thus
$$\lim\limits_{x\to\infty}\frac{1}{x}\int_x^{x+1} f=\lim\limits_{x\to\infty}\frac{f(t(x))}{x}=l\tag{1}$$
The question is how to get from $(1)$ to
$$\lim\limits_{x\to\infty}\frac{f(x)}{x}=l\tag{2}$$
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.
Best Answer
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.