# For a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ what does $\lim_{x\rightarrow +\infty} f'(x)=1$ imply? (TIFR GS $2014$)

Question is :

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

Options:

• $f$ 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.

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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 $f(x) = x$. –  fpqc Dec 8 '13 at 14:18
@Benja : I could not believe i missed this.... :( –  Praphulla Koushik Dec 8 '13 at 14:19

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.)

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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.

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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 –  fpqc 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 –  fpqc Dec 8 '13 at 14:29