TITLE: Limit of the derivative of a function
QUESTION [8 upvotes]: $f(x)$ is a differentiable function on the real line such that $ \lim_{x\to \infty } f(x) =1 $  and $ \lim_{x\to \infty } f'(x) = s $ .Then

$s$ should be $0$
$s$ need not be $0$ but $|s| < 1$
$s > 1$
$s < -1$

Because $f(x)$ is bounded need not mean it can neither be increasing nor decreasing.So the derivative needs not be $0$.

REPLY [3 votes]: Since
$$\lim_{x\to\infty}f'(x)=s$$
That means for all $\epsilon>0$, there exists $R$ and $\delta>0$ s.t.
$$\bigg|\frac{f(x+h)-f(x)}{h}-s\bigg|<\epsilon$$
whenever $x>R$ and $h<\delta$.
In other words,
$$s-\epsilon<\frac{f(x+h)-f(x)}{h}<s+\epsilon$$
Now suppose $s\ne 0$, then set $\epsilon<|s|/2$ and $h=\delta/2$, we have
$$\bigg|f(x+\delta/2)-f(x)\bigg|>\frac{|s|\delta}{4}$$
However, since
$$\lim_{x\to\infty}f(x)=1$$
we know that there exist $R'$ so that
$$\bigg|f(x)-1\bigg|<\frac{|s|\delta}{8}$$
whenever $x>R'$.
Now for $x>\max(R,R')$, we have
$$\bigg|f(x+\delta/2)-f(x)\bigg| \le |f(x+\delta/2)-1|+|f(x)-1|<\frac{|s|\delta}{4}$$
which is a contradiction.
Hence $s=0$.