Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question is :

For a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ what does $\lim_{x\rightarrow +\infty} f'(x)=1$ imply?


  • $f$ is bounded
  • $f$ is increasing
  • $f$ is unbounded
  • $f'$ is bounded

I could not find counter examples but then, I strongly feel $\lim_{x\rightarrow +\infty} f'(x)=1$ would just imply that $f'(x)$ is bounded.

I do not have much idea why would other three are false.

I would be thankful if someone can suggest me some hints.

Thank You.

share|cite|improve this question
If the limit is $1$ then your function is always growing when you move further in $x$ direction. – Tomás Dec 8 '13 at 14:09
but only you can ensure that it is growing from a $ N>0$ large. – Luis Valerin Dec 8 '13 at 14:14
This is enoguh to guarantee that $f$ is unbounded @LuisValerin – Tomás Dec 8 '13 at 14:14
(1) is out upon considering $f(x) = x$. – user38268 Dec 8 '13 at 14:18
@Benja : I could not believe i missed this.... :( – Praphulla Koushik Dec 8 '13 at 14:19
up vote 5 down vote accepted

In fact $f^\prime(x)$ need not be bounded: you can consider if you like $f(x) = x + e^{-x}$; for negative $x$, $|f^\prime(x)|$ can be as large as you like. This function also shows that $f$ can be decreasing on an interval (for instance when $x$ is large and negative).

You can show that, in fact, $f$ must be unbounded, for instance by the fundamental theorem of calculus: for all sufficiently large $a$ and $b>a$, one has $\displaystyle \frac{1}{2}(b-a) = \int_a^b \frac{1}{2}\ dx < \int_a^b f^\prime(x)\ dx = f(b) - f(a)$, so $f(b)$ must grow without bound. (This of course also shows that $f$ is not necessarily bounded.)

share|cite|improve this answer
Oh. yes.. Thank you :) – Praphulla Koushik Dec 8 '13 at 14:19

The only possibility is (3). We eliminate (1),(2),(4) as follows. (1) is out upon considering $f(x) = x$. Also, (4) and (2) are both simultaneously eliminated upon considering the function

$$f(x) = \begin{cases} \arctan x + x, & \text{if $x \geq 0$}\\ x- x^2, & \text{if $ x \leq 0$}.\end{cases}$$

It's derivative at infinity is the same as calculating $\lim_{x \to \infty} (1/(1+x^2) + 1) = 1$. Also, it's non-increasing on $(-1,-1/2)$. For how I came up with the example for (4), well I drew a picture :D

Also let us see why $f$ has to be unbounded. Fix $a$ sufficiently large so that for any $c > a$, $f'(c) \in [1/2,3/2]$. We can do this because $\lim_{x\to \infty} f'(x) = 1$. Then for any $b > a$, the mean value theorem shows there is $d \in [a,b]$ so that

$$\frac{f(b) - f(a)}{b-a} = f'(d).$$

Now the right hand side is at least $1/2$ and so $f(b) \geq 0.5(b-a) + f(a)$. Let $b$ tend to infinity and we see that $f$ is unbounded.

share|cite|improve this answer
I like tour example for $4$.. Thank you :) – Praphulla Koushik Dec 8 '13 at 14:22
@PraphullaKoushik I have modified (4) to give a counterexample for (2) as well :D – user38268 Dec 8 '13 at 14:24
oh yes.... this is quite beautiful :D – Praphulla Koushik Dec 8 '13 at 14:25
@PraphullaKoushik I have a proof of (2) as well :D – user38268 Dec 8 '13 at 14:29
How did you draw the picture @user 38268 – learnmore Nov 15 '14 at 5:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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