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.