Some time ago, I asked this here. A restricted form of the second question could be this:

If $f$ is a function with continuous first derivative in $\mathbb{R}$ and such that $$\lim_{x\to \infty} f'(x) =a,$$ with $a\gt 0$, then $$\lim_{x\to\infty}f(x)=\infty.$$

To prove it, I tried this:

There exist $x_0\in \mathbb{R}$ such that for $x\geq x_0$, $$f'(x)\gt \frac{a}{2}.$$ There exist $\delta_0\gt 0$ such that for $x_0\lt x\leq x_0+ \delta_0$ $$\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&\gt -\frac{a}{4}\\ \frac{f(x)-f(x_0)}{x-x_0}&\gt f'(x_0)-\frac{a}{4}\\ &\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\ f(x)-f(x_0)&\gt \frac{a}{4}(x-x_0)\end{align*}.$$ We can assume that $\delta_0\geq 1$. If $\delta_0 \lt 1$, then $x_0+2-\delta_0\gt x_0$ and then $$f'(x_0+2-\delta_0)\gt \frac{a}{2}.$$ Now, there exist $\delta\gt 0$ such that for $x_0+2-\delta_0\lt x\leq x_0+2-\delta_0+\delta$ $$f(x)-f(x_0+2-\delta_0)\gt \frac{a}{4}(x-(x_0+2-\delta_0))= \frac{a}{4}(x-x_0-(2-\delta_0))\gt \frac{a}{4}(x-x_0).$$ It is clear that $x\in (x_0,x_0+2-\delta_0+\delta]$ and $2-\delta_0+\delta\geq 1$.

Therefore, we can take $x_1=x_0+1$. Then $f'(x_1)\gt a/2$ and then there exist $\delta_1\geq 1$ such that for $x_1\lt x\leq x_1+\delta_1$ $$f(x)-f(x_1)\gt \frac{a}{4}(x-x_1).$$ Take $x_2=x_1+1$ and so on. If $f$ is bounded, $(f(x_n))_{n\in \mathbb{N}}$ is a increasing bounded sequence and therefore it has a convergent subsequence. Thus, this implies that the sequence $(x_n)$: $$x_{n+1}=x_n+1,$$ have a Cauchy's subsequence and that is a contradiction. Therefore $\lim_{x\to \infty} f(x)=\infty$.

I want to know if this is correct, and if there is a simpler way to prove this. Thanks.

  • $\begingroup$ Intuitively (and I am being very inaccurate here) if you think about $f=\int f'$ then the fact that approaching infinity you have a positive value means the area under the curve of $f'$ approaches infinity, which in turns means $f$ approaches infinity. $\endgroup$
    – Asaf Karagila
    Jul 15, 2011 at 9:58
  • $\begingroup$ What about my intent? I'm just curious because in my proof, I don't use the mean value theorem. I know, by the answers, that is simpler with t.m.t., but, since I wrote, I want to know. $\endgroup$
    – leo
    Jul 15, 2011 at 11:31
  • $\begingroup$ I would accept your proof, but it looks like you use the mean value theorem in disguise in your sentence $f(x)-f(x_1)\gt \frac{a}{2}(x-x_1).$ $\endgroup$ Jul 15, 2011 at 12:43
  • $\begingroup$ @leo: I have read the proof. Have a problem digesting the part of the proof that asserts that we can take $\delta_0 \ge 1$. You tried the jump from local (there exists $\delta_0$) to global (we can take $\delta_0 \ge 1$) and I think it didn't work. In the long displayed line just before "it is clear that" the $x$ have the wrong range, they are not near $x_0$. Can be fixed, by using MVT or something a little weaker. The rest of the argument is fine. $\endgroup$ Jul 15, 2011 at 12:58
  • $\begingroup$ @Ross Millikan: You're right, but it's possible obtain something like $f(x)-f(x_1)\gt \frac{a}{4}(x-x_1)$. I'll fix it. $\endgroup$
    – leo
    Jul 15, 2011 at 19:40

6 Answers 6


Since you assume that $f$ has first derivative continuous on $\mathbb{R}$, there is an alternative to the Mean-value theorem, namely the Fundamental theorem of calculus (the former is very useful for cases when $f'$ is not assumed continuous). So, from $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. Since $f'$ is assumed continuous, we can apply the the Fundamental theorem of calculus to obtain $$ f(x) - f(M) = \int_M^x {f'(u)\,du} \ge \int_M^x {\frac{a}{2}\,du} = \frac{a}{2}(x - M), $$ for any $x > M$. Since the right-hand side goes to $\infty$ as $x \to \infty$ (note that $M$ is fixed), so does $f(x) - f(M)$. Hence $ \lim _{x \to \infty } f(x) = \infty $.

EDIT: The above was just to give an alternative to the the Mean-value theorem (used in Amitesh Datta's answer); its analog for the Mean-value theorem is as follows.

From $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. For any $x > M$, since $f$ is continuous on the closed interval $[M,x]$ and differentiable on the open interval $(M,x)$, by the Mean-value theorem there exists $c = c(M,x) \in (M,x)$ such that $$ f(x)-f(M)=f'(c)(x-M). $$ Since $c \geq M$, $f'(c) > a/2 $, and hence $$ f(x)-f(M) \geq \frac{a}{2}(x - M). $$ The conclusion is as before.

Note: As already indicated above, the proof using the Mean-value theorem has the advantage that it works for any differentiable function $f: \mathbb{R} \to \mathbb{R}$.

  • $\begingroup$ +1 Nice answer! I think it is great to show alternative methods for solving a particular problem as you have done. Also, your method using the mean value theorem is much cleaner than mine since there is a complicated constant in my answer (but the idea is the same) ... $\endgroup$ Jul 15, 2011 at 12:43
  • $\begingroup$ @Amitesh: Thanks! (+1 to you too, for you provided a proof using MVT first.) $\endgroup$
    – Shai Covo
    Jul 15, 2011 at 12:52
  • $\begingroup$ This answer is nice too (as is Bill's, by the way). I decided to rephrase it using more physical language in my own answer below... $\endgroup$ Jul 16, 2011 at 8:15
  • $\begingroup$ This is an elegant answer. $\endgroup$
    – leo
    Jul 16, 2011 at 12:43
  • $\begingroup$ @Pete and leo: Thank you. $\endgroup$
    – Shai Covo
    Jul 17, 2011 at 4:11

HINT $\ \ $ Writing $\rm\ f\ =\ (f/x)\ x\ \ $ yields a determinate limit (using L'Hôpital's rule), namely

$$\rm\displaystyle\quad \lim_{x\to\infty}\ f\ =\ \lim_{x\to\infty} \dfrac{f}x\ \lim_{x\to\infty}\ x\ =\ \lim_{x\to\infty}\dfrac{f\:\:'}{1\:\ }\ \lim_{x\to\infty}\ x\ =\ a\ \cdot \infty\ =\ \infty\quad by\quad a > 0$$

NOTE $\ $ Below is said L'Hospital's rule, from Rudin's $\:$ Principles of Mathematical Analysis, $\:$ 1976. Above I use the general form $(15)$ of the $\rm \infty/\infty$ rule. It requires only that the denominator $\to\infty\:.$ For further such generalizations see the Monthly papers cited here.

enter image description here

  • $\begingroup$ How do you justify the first equation? $\endgroup$
    – t.b.
    Jul 15, 2011 at 16:26
  • $\begingroup$ @Theo Could you please elaborate. $\endgroup$ Jul 15, 2011 at 16:39
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    $\begingroup$ Ah, right. I was definitely being silly. Thanks for setting me straight. $\endgroup$
    – t.b.
    Jul 15, 2011 at 17:19
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    $\begingroup$ @Bill: +1, Rudin really does have a nice improvement on the standard L'Hopital's Rule. I think though that there may be a reason that this does not appear in freshman calculus texts: the idea of an indeterminate form is such a tricky one for students at this level that having a version of L'Hopital's Rule that applies to many determinate forms may be asking for trouble. (In particular the other answer where you applied this L'Hopital's Rule did not make me a fan: I don't want a calculus student to apply a named theorem to conclude that $\frac{0}{\infty} = 0$...) $\endgroup$ Jul 16, 2011 at 8:27
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    $\begingroup$ @Bill: okay. From one experienced calculus teacher to another: students often misapply L'Hopital's rule and get confused by it. So one has to make careful choices about what to say and what not to say. So e.g. I do give Rudin's version of L'Hopital in my "honors calculus notes" math.uga.edu/~pete/2400full.pdf. (These notes were written after my last comment.) But I think I did not mention it in the course itself. $\endgroup$ Feb 4, 2014 at 19:39

There is a simpler way to prove this. (I apologize but I have not looked at your solution; perhaps another user will comment in this regard.) Let $M$ be a positive integer. Choose a positive integer $N$ such that $x>N$ implies $f'(x)>\frac{a}{2}$. If $y>\text{max}\{N,\frac{2(M-f(N))}{a}+N\}$, then $y>N$ and the mean value theorem implies that $f(y)-f(N)=f'(x_y)(y-N)$ for some real number $N<x_y<y$. Furthermore, $f(y)=f(N)+f'(x_y)(y-N)>f(N)+\frac{a(y-N)}{2}>M$ for all such $y$. Since $M$ was arbitrary, it follows that $\lim_{x\to\infty} f(x)=\infty$.

  • $\begingroup$ +1: I like this solution best because it is relatively short and direct and because it openly uses MVT. (The other solutions use results which require MVT to prove. There is of course nothing wrong with that but yours feels more "honest" somehow.) $\endgroup$ Jul 16, 2011 at 6:08
  • $\begingroup$ @Pete: Dear Pete, thanks very much for your kind words! $\endgroup$ Jul 16, 2011 at 7:40

After reading some of the other answers, I thought it might be helpful to rephrase Shai Covo's answer in more physical language: suppose that instead of $y = f(x)$ we have $x = x(t)$, the position function of a particle at time $t$. (Of course it makes no mathematical difference what we call the variables...)

The hypothesis is that the velocity function $x'(t)$ approaches a limiting value $v_0 > 0$ as $t$ approaches infinity. Therefore there must exist a $t_0 \in \mathbb{R}$ such that for all $t \geq t_0$, $v(t) \geq \frac{v_0}{2} > 0$. Let us therefore compare the position function $x(t)$ to the following one:

$$y(t) = \frac{v_0}{2}\left(t-t_0\right) + x(t_0).$$

We observe:

1) $y(t)$ is a linear function with positive slope, so certainly $\lim_{t \rightarrow \infty} y(t) = \infty$.

2) We have $x(t_0) = y(t_0)$ and $x'(t) \geq y'(t)$ for all $t \geq t_0$.

Thus on the time interval $[t_0,\infty)$, the particle $x = x(t)$ starts out at the same position as the particle $y = y(t)$ and for all times $t$ the instantaneous velocity of the first particle is always greater than or equal to that of the second particle. Therefore physical intuition tells us that we must have $x(t) \geq y(t)$ for all $t \geq t_0$, and hence $\lim_{t \rightarrow \infty} x(t) = \infty$.

How do we make this rigorous? Replacing $x(t)$ and $y(t)$ by $x(t) - y(t)$ and $0$, what we need to show is that a function which is non-negative at a point $t_0$ and has everywhere non-negative derivative is itself non-negative for all $t \geq t_0$. But now we can take our pick of mathematical justifications for this: if we assume that the derivative is continuous (as the OP does) then $x(t) = \int_{t_0}^t x'(t) dt$ and the integral of a non-negative function is certainly non-negative, being a limit of non-negative Riemann sums. Or indeed, if we assume that there exists $t_1 > t_0$ such that $x(t_1) < x(t_0)$ and apply the Mean Value Theorem, then we get

$$x(t_1) - x(t_0) = x'(c) (t_1-t_0).$$

The left hand side is negative and the right hand side is non-negative: contradiction!

In summary, reasoning physically in a mathematically careful way reduces us to a statement which is easy to prove mathematically.

As a side remark, some educators have suggested that this special case of the Mean Value Theorem (called the Mean Value Inequality, Increasing Function Theorem, and so forth) is indeed more intuitive than the full Mean Value Theorem and have proposed developing calculus around it. In my experiences teaching calculus I have found it helpful to point out explicitly this special case of MVT, but that's about it: the full MVT is also (more) useful and not much more bitter a pill...

  • $\begingroup$ This a nice answer. $\endgroup$
    – leo
    Jul 16, 2011 at 12:41

Since $\lim_{x\rightarrow \infty}f'(x) = a > 0$, there exists some $x_0 > 0$ such that $f'(x) \geq \frac{a}{2}$ for every $x >x_0$. For any $\alpha > x_0$, consider the function $\Phi: [x_0, \alpha] \rightarrow \mathbb{R}$ defined by $\Phi(x) = f(x)$ for every $x \in [x_0, \alpha]$. This function is continuous on $[x_0, \alpha]$ and differentiable on $(x_0, \alpha)$ for every $\alpha > x_0$. Therefore, it satisfies all the hypotheses of the Lagrange Mean Value Theorem. Hence, there exists $\gamma(x_0, \alpha) \in (x_0, \alpha)$ such that

$$f(\alpha) - f(x_0) = f'(\gamma(x_0, \alpha))(\alpha - x_0),$$ and this implies that $f(\alpha)>f(x_0)+\frac{a}{2}(x-x_0)$ for every $\alpha>x_0$. But note that $a>0$, therefore $\lim_{\alpha\rightarrow \infty}f(x_0)+\frac{a}{2}(x-x_0)=\infty$. This shows that $\lim_{\alpha\rightarrow \infty}f(\alpha)=\infty$.

I hope this helps. If you spot any gaps in my proof I would be grateful to hear any constructive criticisms.


I will try to prove is in a different way which can be much simpler - using visualization.

Imagine how will a function look if it has a constant, positive slope -
A straight line, with a positive angle with the positive x axis.
Although this can be imagined, I am attaching a simple pic -
entera image description here (Plot of our imaginative function - $f(x)$ vs $x$)

As per the situation in the question, for $f(x)$ the slope exists (and is finite) at all points, so it means that the function is continuous. Since the slope is also constant at $\infty$, $f$ has to be linear at $\infty$. Thus, the graph of the function should be similar to the above graph.
(assume the value of x to be as large as you can imagine.)

Hence, putting the above situation mathematically, we have,

If $\lim_{x\to \infty}\ f'(x) =a\qquad(with\ a>0)$

then $\lim_{x\to\infty}\ f(x)=\infty.$

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    $\begingroup$ @Guanidene: but the function $f$ is not assumed to have constant positive slope, only that the limit of the slope of the tangent line exists and is positive. Your solution amounts to assuming that $f$ is linear (and exhibiting a straight line for good measure -- I don't want to speak for others, but actually I have seen these before!). Also as a past and future calculus teacher I find your comment rather rude: I don't like this answer because it makes unwarranted assumptions and is thus incorrect. Three other correct answers have already been given. $\endgroup$ Jul 16, 2011 at 6:06
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    $\begingroup$ @Guanidene: $\infty$ is not a real number, so your arguments are not rigorous. Also the graph that you have displayed is labelled with finite values of $x$, not infinite values of $x$. I think it could be possible to make your argument rigorous by using non-standard analysis, but as it stands it is is simply not a proof at all. $\endgroup$ Jul 16, 2011 at 6:31
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    $\begingroup$ @Guanidene: your last two comments contradict each other. Moreover, your attitude seems combative -- you don't seem interested in why several professional mathematicians think your argument is incorrect -- so I am giving up for now. $\endgroup$ Jul 16, 2011 at 6:37
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    $\begingroup$ @Guanidene: I don't understand why you're assuming the OP does not want a proof (he asks for one!), but there's no point in arguing about it further since he can accept your answer if it indeed suits his purposes. $\endgroup$ Jul 16, 2011 at 7:43
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    $\begingroup$ I'm able to proof such simple question. I don't do with M.V.T. because I was doing things like I did in my own answer. But the other answers are totally clear for me. It seems, as in your answer, you do wrong assumptions. Thank you if you want to help me. But... Is not that the teachers (or professionals) don't accept proofs by picture, just simply, in math, this does not exist. $\endgroup$
    – leo
    Jul 16, 2011 at 13:10

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