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.
 A: 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.)
A: 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.
