In the book of The elements of Real Analysis, by Bartle, at page 220, it is asked to show that

If $f(x)\to a$ and $f'(x) \to b$ as $x\to +\infty$, then $b = 0$

However, I'm having trouble showing the result.

By our assumption, for a given $\epsilon > 0$, $\exists M \in \mathbb{R}$ and $\delta >0 $ such that $\forall x > M$, $$|f(x) - a| < \epsilon,$$ and $$|\frac{f(x+h) - f(x)}{h} - b| < \epsilon$$ for $0 < |h| < \delta.$

However, (by observing $x+h > x > M$), I get $$-2\epsilon / h - b<|f'(x) - b| < 2\epsilon / h - b,$$ which does not let me do anything because of that $h$ in the denominator of $\epsilon$. So my question is, how can we prove this result?

  • 1
    $\begingroup$ If $b>0$ then $f$ eventually lies above a line of positive slope $b/2$. If $b<0$ then $f$ eventually lies below a line of negative slope $b/2$. In either case, $f$ tends to $\pm\infty$ as $x \to \infty$. Thus the only way $f$ can tend to $a$ is if $b=0$. $\endgroup$ – nullUser Dec 21 '17 at 6:51
  • $\begingroup$ @nullUser Thanks for you answer. However, in somewhat similar whats, I can see the result intuitively, too, and what I can't see is that how to deal with those epsilons, so I would really appreciate you could post your answer/comment by supporting it with a rigorous mathematical proof. $\endgroup$ – onurcanbektas Dec 21 '17 at 6:54
  • 3
    $\begingroup$ Mean value theorem is your friend here. Try it on the difference $f(x+1)-f(x)$. $\endgroup$ – Paramanand Singh Dec 21 '17 at 6:57
  • $\begingroup$ To do that, I need to choose $h = 1$, but that would require $\delta > 1$, but for a given $\epsilon > 0$, I cannot guarantee that. $\endgroup$ – onurcanbektas Dec 21 '17 at 7:01
  • 1
    $\begingroup$ Here's a little intuition to go with it. If $f$ tends to a certain fixed number, $a$, as $x$ goes to infinity, we can say that $f'$ goes to $0$ at infinity since if it didn't there would be no way that $f \to a$. $\endgroup$ – ThisIsNotAnId Dec 21 '17 at 7:02

Correct me if wrong :

1)$ \lim_{x \rightarrow \infty} f(x) =a$:

Let $\epsilon$ be given .

There exists an $M$, real, such that for $x\ge M $

$|f(x)-a| \lt \epsilon.$

2) MVT:

Consider $h \gt 0$, $h$ fixed.

$hf'(t) = f(x+h) - f(x)$, with

$x \lt t \lt x+h$.

Let $x \gt M$.

$h|f'(t)| = |f(x+h) -f(x)| =$

$ |(f(x+h) -a) -(f(x) -a)| \le$

$|f(x+h) -a| +|f(x)-a| \lt 2\epsilon.$

Recall : $x \lt t\lt x+h:$

$ x \rightarrow \infty$ implies $t \rightarrow \infty$ :

$h \lim_{t \rightarrow \infty} |f'(t)|= h|b| \le 2\epsilon.$

This implies;

$\lim_{t \rightarrow \infty}|f'(t)|=|b|= 0.$

Note: If $|b| \not=0$ we get a contradiction by choosing $h$, an independent parameter, sufficiently large.

  • $\begingroup$ İt needs to be $x \lt t \lt x+h$ $\endgroup$ – onurcanbektas Dec 21 '17 at 9:45
  • $\begingroup$ Even though the rest of the proof is OK, and it is constructed in the way that I was looking, the organisation of the lines makes the answer hard to read. $\endgroup$ – onurcanbektas Dec 21 '17 at 9:50
  • $\begingroup$ Note that, if you do not organise your answer, I cannot accept it. $\endgroup$ – onurcanbektas Dec 21 '17 at 10:24
  • $\begingroup$ Thanks. The typo is fixed. $\endgroup$ – Peter Szilas Dec 21 '17 at 14:53

Just to expand the diversity of the answers, I'm going to post the proof suggested by @ParamanandSingh

Let for a fixed $x\in \mathbb{R}$, consider the interval $(x, x+1)$. By MVT, $\exists c \in (x, x+1)$ s.t $$\frac{f(x+1) - f(x)}{1} = f'(c).$$

Now if we let $x\to +\infty$, $c\to +\infty$, and the difference $$f(x+1) - f(x) \to 0,$$ hence $$f'(c) \to b\quad as \quad c\to +\infty$$ implies $$b = 0$$


  • 2
    $\begingroup$ It's almost correct. Just write that the LHS $f(x+1)-f(x)$ tends to $a-a=0$ and RHS $f'(c) $ tends to $b$ (given in question). So $b=0$ and you are done. This proof is very standard and available in many good textbooks. $\endgroup$ – Paramanand Singh Dec 21 '17 at 10:00
  • $\begingroup$ @ParamanandSingh Ok, thanks for pointing out. $\endgroup$ – onurcanbektas Dec 21 '17 at 10:07
  • 2
    $\begingroup$ I think you should write $f'(c) \to b$ and hence $b=0$. Note that $f'(c) $ may never be equal to $b$ (a limit is not necessarily a value attained). So that equal sign might raise some doubts. +1 anyway. And if you haven't noticed so far your answer is the shortest and simplest. $\endgroup$ – Paramanand Singh Dec 21 '17 at 10:12
  • $\begingroup$ @ParamanandSingh Yeah, I have noticed :) $\endgroup$ – onurcanbektas Dec 21 '17 at 10:14

If we assume

$f'(x) \to b \; \text{as} \; x \to \infty, \tag 1$

then we have, for any $\epsilon > 0$, a sufficiently large $M \in \Bbb R$ such that

$b - \epsilon < f'(x) < b + \epsilon \; \text{for} \; x \ge M; \tag 2$

suppose that $b > 0$; then we choose $\epsilon$ so small that

$b - \epsilon > 0; \tag 3$


$f(x) - f(M) = \displaystyle \int_M^x f'(s) \; ds \ge \int_M^x (b - \epsilon)\; ds = (b - \epsilon)(x - M), \tag 4$


$f(x) \ge f(M) + (b - \epsilon)(x - M) \to \infty \; \text{as} \; x \to \infty; \tag 5$

likewise, if $b < 0$, we may choose $\epsilon$ such that

$b + \epsilon < 0; \tag 6$

then by an argument similar to the above we have

$f(x) \le f(M) + (b + \epsilon)(x - M) \to -\infty \; \text{as} \; x \to \infty; \tag 7$

in neither case $b > 0$, $b < 0$ does $f(x) \to a$ as $x \to \infty$; thus, we must have

$b = 0. \tag 8$

  • 1
    $\begingroup$ Well, even though I know what is integral, and what does geometrically mean, in the book, we haven't covered that, so it would be really nice if your argument is only based on the algebraic manipulations. $\endgroup$ – onurcanbektas Dec 21 '17 at 8:01
  • 1
    $\begingroup$ @onurcanbektas: well, I had no way of knowing what tools you had at your disposal until I read your comment. I'm not familiar with Bartle's book, an not in a place where I have access to a copy at the moment. $\endgroup$ – Robert Lewis Dec 21 '17 at 8:05
  • 1
    $\begingroup$ Of course, I should have notes those kinds of thing in the question, it is my mistake. $\endgroup$ – onurcanbektas Dec 21 '17 at 8:06
  • 1
    $\begingroup$ @onurcanbektas: The mean value theorem approach suggested by Paramanand Singh might hold some promise, but I'm to sleepy right now to put all the details together. $\endgroup$ – Robert Lewis Dec 21 '17 at 8:07
  • 1
    $\begingroup$ @onurcanbektas: not too bad of a mistake! Anyway, I just went for the mathematical interest of the question since I don't know how your course is set up. Cheers! $\endgroup$ – Robert Lewis Dec 21 '17 at 8:09


As it is pointed out, there is a mistake in this proof, but it can be solved by observing that the difference $f(x+1)- f(x)$ can be arbitrarily small for sufficient large values of $x$.

After getting some ideas, it is clear that easiest method is to use the method of contradiction, so I'm going to post my own proof in here.

WLOG, let $f'(x) > 0$ as $x\to +\infty$, then by the lemma $19.3.a$ in Bartle's book, saying that, $\exists h> 0$ s.t for all y satisfying $x< y < x+h$, we have $f(x)< f(y)$. Then let $f(y) - f(x) = \epsilon'$ and choose $2\epsilon < \epsilon'$ so that $\exists M$ and $\forall x > M$, we have $$|f(x) - a| < \epsilon.$$ Since $y > x > M,$ $|f(y)- a| < \epsilon$, hence $$0 < \epsilon' < |f(y)- f(x)| < 2\epsilon$$, which is a contradiction. QED

  • 1
    $\begingroup$ This isn’t right - your $\epsilon’$ depends on $x,y$ but your $x,y$ depend on $\epsilon’.$ Maybe you can show there is a positive lower bound on $f(x+1)-f(x)$ for sufficiently large $x.$ $\endgroup$ – Dap Dec 21 '17 at 8:11
  • $\begingroup$ @Dap You are right. $\endgroup$ – onurcanbektas Dec 21 '17 at 10:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.