Below is a problem from Chapter 22 "Infinite Sequences" of Spivak's Calculus. I've previously asked another question regarding an incorrect attempt at solving item $b$ of this problem. Now I am interested in the solution manual solution.

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 $[n,n+1]$.

Here is the solution manual solution to part $(b)$

The hypothesis $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$ implies that $a_n$ and $b_n$, the $\inf$ and $\sup$ of $f$ on $[n,n+1]$, satisfy $\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$ and $\lim\limits_{n\to\infty} [b_{n+1}-b_n]=l$.

So by part $(a)$, we have $\lim\limits_{x\to\infty} a_n/n=\lim\limits_{x\to\infty} b_n/n=l$, which implies that $\lim\limits_{x\to\infty} f(x)/x=l$.

How do we conclude

$$\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$$


$$\lim\limits_{n\to\infty} [b_{n+1}-b_n]=l$$


$$\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$$


  • $\begingroup$ @OliverDíaz I am in the process of going through that solution. It seems substantially more complex than this one from the solution manual, however, right? Or is it simply the case that this solution manual solution has omitted a bunch of intermediate complicated steps used to prove the question I am currently asking? $\endgroup$
    – xoux
    Oct 20, 2022 at 7:17
  • $\begingroup$ @OliverDíaz Are you referring to $a_n\leq f(x)\leq f(x+1)\leq b_n$ then $f(x+1)-f(x)\leq b_n-a_n$? $\endgroup$
    – xoux
    Oct 20, 2022 at 7:49
  • 1
    $\begingroup$ We do not have $f(x)\leq f(x+1)$. This $f(x+1)-f(x)\ge0$ is not implied by the existence of $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right).$ $\endgroup$ Oct 20, 2022 at 8:16
  • 1
    $\begingroup$ @AnneBauval: Notice my last comment addressed to you. I corrected what I said. I am not claiming $a_n\leq a_{n+1}\leq b_{n+1}\leq b_n$. $\endgroup$
    – Mittens
    Oct 20, 2022 at 8:16
  • 1
    $\begingroup$ We do not have that "above some value the graph of $f$ starts resembling something linear and increasing". E.g. $f(x)=\sin(2\pi x).$ $\endgroup$ Oct 20, 2022 at 8:53

4 Answers 4


Given that $a_n:=\inf f[n,n+1], b_n=\sup f[n,n+1] .$

Well-definedness: $a_n,b_n$ are well defined since $f$ is continuous and $[n,n+1]$ is a compact set (this is same as closed and bounded in $\mathbb R$).

To prove: $\lim_n (a_{n+1}-a_n)=l=\lim_n (b_{n+1}-b_n),$ which by definition of limit is equivalent to proving that "Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$"

To achieve this, first fix an $\epsilon>0$.

Let's break down rest of the solution in steps:-

Step (1): (Choosing large $x$) Since $\lim_{x\to \infty}(f(x+1)-f(x))=l$, there exists $N\in \mathbb N$ such that $x\ge N\implies -\epsilon/2\lt f(x+1)- f(x)-l\lt \epsilon/2.$

Step(2):(Estimating $a_{n+1}-a_n$ from above) Note that for all $n\ge 2N,$

By definition of infimum, there exists $x_n\in [n,n+1]$ such that $a_n+\epsilon/2\gt f(x_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \le &a_{n+1}-f(x_n)-l+\epsilon/2\\ \le& f(x_n+1)-f(x_n)-l+\epsilon/2\\ \overbrace{\le }^{\text{ By step $(1)$}}& \epsilon/2+\epsilon/2=\epsilon \end{align}$$

Step (3):(Estimating $a_{n+1}-a_n$ from below) Note that for all $n\ge 2N,$

By definition of infimum, there exists $y_n\in [n+1,n+2]$ such that $a_{n+1}+\epsilon/2\gt f(y_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \ge & f(y_n)-a_n-l-\epsilon/2\\ \ge& f(y_n)-f(y_n-1)-l-\epsilon/2\\ \overbrace{\ge }^{\text{ By step $(1)$ again}}& -\epsilon/2-\epsilon/2=-\epsilon \end{align}$$

It follows by step (2) and (3) that: Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge 2N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$ This proves the statement for $(a_{n+1}-a_n)$. The statement is similar for $(b_{n+1}-b_n)$, which I leave up to you.


Applying the general property $$\inf_{x\in A} (a(x)+b(x))\ge\inf_{x\in A} a(x)+\inf_{x\in A} b(x)$$ to $A=[n,n+1]$, $a(x)=f(x+1)-f(x)$ and $b=f$, we get: $$a_{n+1}\ge\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)+a_n.$$ Similarly, $$a_n\ge\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right)+a_{n+1}.$$ These two inequalities are summarized in $$\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)\le a_{n+1}-a_n\le-\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right).$$ From this, using the equality $-\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right)=\sup_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)$ and the hypothesis $\lim\limits_{x\to\infty}\left(f(x)-f(x+1)\right)=l$, we conclude: $$\lim\limits_{n\to\infty}\left(a_{n+1}-a_n\right)=l.$$ The similar result for $b_n$ may be proved the same way, or deduced from the result for $a_n$ by changing $f$ to $-f.$

  • 1
    $\begingroup$ +1 this is the cleanest way $\endgroup$
    – FShrike
    Oct 20, 2022 at 8:02
  • $\begingroup$ Would you mind if I edit your answer to number your equations? In the inequality $a_n\ge\inf_{x\in[n,n+1]}\left(f(x)-f(x+1)\right)+a_{n+1}.$ shouldn't it be a $\leq$ sign? $\endgroup$
    – xoux
    Oct 20, 2022 at 8:13
  • $\begingroup$ Please don't change. $\endgroup$ Oct 20, 2022 at 8:21
  • $\begingroup$ The inequality $a_{n+1}\ge\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)+a_n.$ isn't obvious to me. $\inf_{x\in[n,n+1]}\left(f(x+1)-f(x)\right)$ means that we take every point $x$ in $[n,n+1]$ and compare $f(x)$ to $f(x+1)$ and take the smallest, call it $\gamma$. Then we are saying that the difference between $a_n$ and $a_{n+1}$, which is the difference between the smallest value of $f$ in $[n,n+1]$ and in $[n+1,n+2]$, is larger than $\gamma$. Is this obvious? I can't come up with a counterexample, but it doesn't follow crystal clear from the first line. $\endgroup$
    – xoux
    Oct 20, 2022 at 8:42

Perhaps this answer does not fit, because I prefer not to follow Spivak's suggestion.

Let $g(x)=f(x)-lx.$ Then $\displaystyle\lim_{x\to \infty} [g(x+1)-g(x)]=0.$ We are done once we get $$\lim_{x\to \infty}{g(x)\over x}\overset{?}{=}0\quad (*)$$ Fix $\varepsilon >0.$ There exists a number $x_0\ge 1$ such that $$|g(x+1)-g(x)|<\varepsilon,\qquad x+1>x_0$$ For $x>x_0$ let $n_x$ denote the smallest nonnegative integer such that $x-n_x\le x_0.$ Hence $x-n_x+1>x_0.$ Thus $$ n_x<x-x_0+1\le x$$ We have $$\displaylines{g(x)=[g(x)-g(x-1)]+[g(x-1)-g(x-2)]\\ \qquad \qquad \qquad +\ldots +[g(x-n_x+1)-g(x-n_x)]+g(x-n_x)}$$ Therefore $$|g(x)|\le n_x\varepsilon +\max_{0\le t \le x_0}|g(t)|$$ Hence for $\displaystyle c:=\max_{0\le t\le x_0}|g(t)|$ we have $${|g(x)|\over x}\le {n_x\over x}\varepsilon+{1\over x}c\le \varepsilon +{1\over x}c,\quad x>x_0-1$$ Finally $${|g(x)|\over x}<2\varepsilon,\qquad x>x_0-1,\ x>{c\over \varepsilon}$$ This completes the proof of $(*).$

  • $\begingroup$ (+1) a similar solution was posted before in a link tied to this problem, but your posting is much clearer. $\endgroup$
    – Mittens
    Oct 20, 2022 at 23:40

Part(b) follows from part (a).

Here I discussed two solutions: one I obtained by following the link given in Spivak's book (solution 1), and an explanation of the solution in the user's manual described in the OP (solution 2). At the end I also add simple solution to part (a) of the problem in the OP.

Solution 1: First notice that \begin{align} \frac{f(x)}{x}=\frac{f(x)-f(\lfloor x\rfloor)}{x} +\frac{\lfloor x\rfloor}{x}+ \frac{f(\lfloor x\rfloor)}{\lfloor x\rfloor} \end{align} where $\lfloor x\rfloor$ is the largest less or equal to $x$. Since $\lim_{x\rightarrow\infty}f(x+1)-f(x)=\ell$, $f(\lfloor x\rfloor)/\lfloor x\rfloor \xrightarrow{x\rightarrow\infty}\ell$ by part (a). Thus, it is enough to show that \begin{align} \lim_{x\rightarrow\infty}\frac{f(x)-f(\lfloor x\rfloor)}{x}=0\tag{0}\label{zero} \end{align} Being that that $f$ is bounded in any compact interval, for any $n\in\mathbb{N}$ $\alpha_n=\inf_{x\in[n,n+1]}f(x)$ and $\beta_n=\sup_{x\in[n,n+1]}f(x)$ are finite. Then, $|f(x)-f(\lfloor x\rfloor)|\leq \beta_n-\alpha_n$ for $n=\lfloor x\rfloor$.

Give $\varepsilon>0$, there is $N$ large enough such that $x\geq N$ implies \begin{align} f(x)+\ell-\varepsilon<&f(x+1)<f(x)+\ell+\varepsilon\tag{1}\label{one}\\ f(x+1) -\ell-\varepsilon &< f(x)<f(x+1)-\ell+\varepsilon\tag{2}\label{two}& \end{align} From \eqref{one} and \eqref{two} we obtain \begin{align} \alpha_n+\ell-\varepsilon&\leq \alpha_{n+1}\leq \beta_{n+1}\leq \beta_n+\ell+\varepsilon,\tag{3}\label{three}\\ \alpha_{n+1}-\ell-\varepsilon&\leq \alpha_n\leq \beta_n\leq \beta_{n+1}-\ell+\varepsilon,\tag{4}\label{four} \end{align} for all $n\geq N$ and so, \begin{align} \alpha_n-\beta_n-2\varepsilon\leq \beta_{n+1}-\alpha_{n+1}\leq \beta_n-\alpha_n+2\varepsilon\tag{5}\label{five} \end{align} Iterating this inequality yields

\begin{align} \beta_N-\alpha_N - (n-N+1)2\varepsilon\leq \beta_{n+1}-\alpha_{n+1}\leq (\beta_N-\alpha_N)+ (n-N+1)2\varepsilon\tag{6}\label{six} \end{align} It follows that for all $x\geq N+1$, $$\frac{|f(x)-f(\lfloor x\rfloor)|}{x}\leq \frac{\beta_N-\alpha_N}{x}+\frac{\lfloor x\rfloor -N}{x}2\varepsilon$$ This means that $\limsup_{x\rightarrow\infty}\frac{|f(x)-f(\lfloor x\rfloor)|}{x}\leq2\varepsilon$ for any $\varepsilon>0$, and \eqref{zero} follows.

Solution 2: As for what is in the solution's manual, notice that from \eqref{three} and \eqref{four} \begin{align} \ell-\varepsilon&\leq \alpha_{n+1}-\alpha_n\leq \ell+\varepsilon\\ \ell-\varepsilon&\leq \beta_{n+1}-\beta_n\leq \ell+\varepsilon \end{align} for all $n\geq N$. This implies that $$\lim_n(\beta_{n+1}-\beta_n)=\ell=\lim_n(\alpha_{n+1}-\alpha_n)$$ By part (a) of the problem $$\lim\frac{\alpha_n}{n}=\ell=\lim_n\frac{\beta_n}{n}$$ wich also implies that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\ell$.

Part(a) follows by taking Cesaro sums: since $a_{n+1}-a_n\xrightarrow{n\rightarrow\infty}\ell$, $\ell=\lim_n\frac1n\sum^{n-1}_{k=0}(a_{k+1}-a_k)=\lim_n\frac{a_n-a_0}{n}=\ell$.

  • $\begingroup$ This is a useful answer in the context of my previous question that regarded general solutions and I am laboring to go through it. However, the current question is very specific: for the specific solution from the solution manual, how does one go from $\lim\limits_{x\to\infty} [f(x+1)-f(x)]=l$ to $\lim\limits_{n\to\infty} [a_{n+1}-a_n]=l$ and $\lim\limits_{n\to\infty} [b_{n+1}-b_n]=l$? $\endgroup$
    – xoux
    Oct 20, 2022 at 7:25
  • $\begingroup$ Why does part $(a)$ make it enough to show $(0)$? $\endgroup$
    – xoux
    Oct 20, 2022 at 8:50
  • $\begingroup$ @evianpring: Becase (1) taking the limit over the integers $\lim_n f(n)/n=\ell$ by virtue of $\lim_n(f(n+1)-f(n))=\ell$; (2) taking the limit over reals, $\lim_{x\rightarrow\infty}x/\lfloor x\rfloor =1$. I edited my solution for the last time to address also the solution in your manual using my main arguments. Hope this helps. $\endgroup$
    – Mittens
    Oct 20, 2022 at 9:19

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