Spivak, Ch. 22 "Infinite Sequences", Problem 17b: $f$ cont., $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$. Prove $\lim\limits_{x\to\infty} [f(x)/x]=l$ 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.
 A: 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.
A: 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 $f(x)/x\rightarrow l$.
