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.

• Why $\int_{1}^{x+1}-\int_{0}^{1}$ would become $\int_{x}^{x+1}-\int_{0}^{1}$? Oct 20, 2022 at 0:06
• @user284331 there was a typo. It should be $\int_1^{x+1}-\int_0^x$ that becomes $\int_x^{x+1}-\int_0^1$. Oct 20, 2022 at 0:11
• Oct 20, 2022 at 1:49

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 $$f(x)/x\rightarrow l$$.