If $\lim_{x \rightarrow \infty} f(x)$ is finite, is it true that $ \lim_{x \rightarrow \infty} f'(x) = 0$? Does finite $\lim_{x \rightarrow \infty} f(x)$ imply that $\lim_{x \rightarrow \infty} f'(x) = 0$? If not, could you provide a counterexample?
It's obvious for constant function. But what about others?
 A: $$\lim_{x\to \infty} f'(x) = \lim_{x\to \infty} \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
Your claim would be true if it were allowed to switch the order of the two limits, which is not possible for all functions.
A: The cities of MinusOneVille and PlusOneVille are connected by a straight railway. One morning, two identical trains leave each city, in the direction of the other city. A nerdy math fly sits at the front of one train, and since it is in a good mood this morning, s/he decides to fly back and forth between the trains and enjoy the view in both directions.
Now, the trains don't want to hit each other, so they gradually decelerate, so that every hour the distance between them is halved. The fly, however, flies at constant speed in every direction (and decelerates and accelerates more and more quickly when arriving at each train, so that it spends most of every run in flight between the trains).
Now, the limit of the fly's position as time goes by is halfway between the two cities (mark it 0), but its speed does not tend to 0, as most of the time it's flying between the trains at its good-mood airspeed.
$x$ is the time, $f(x)$ is the fly's position, $f'(x)$ is its speed - so $\lim_{x \rightarrow\infty}{f(x)} = 0$ but $\lim_{x\rightarrow\infty}{f'(x)} \neq 0$.
A: Simple counterexample: $f(x) = \frac{\sin x^2}{x}$.
UPDATE: It may seem that such an answer is an unexplicable lucky guess, but it is not. I strongly suggest looking at Brian M. Scott's answer to see why. His answer reveals exactly the reasoning that should first happen in one's head. I started thinking along the same lines, and then I just replaced those triangular bumps with $\sin x^2$ that oscillates more and more quickly as $x$ goes to infinity.
A: If you don’t see an example using known functions, you can think about it this way.
Start with a piecewise linear function $f$ whose graph lies along the $x$-axis except in the intervals $\left[n-\frac1{n^2},n+\frac1{n^2}\right]$ for $2\le n\in\Bbb Z$; on $\left[n-\frac1{n^2},n\right]$ it rises linearly from $0$ to $\frac1n$, and on $\left[n,n+\frac1{n^2}\right]$ it falls linearly to $0$. Clearly $\lim\limits_{n\to\infty}f(x)=0$. This function isn’t differentiable at the points $n$ and $n\pm\frac1{n^2}$ for $n\ge 2$, but it’s clearly possible to smooth out the corners. Note that the slope of the graph on $\left[n-\frac1{n^2},n\right]$ is $n$; if you smooth out the corners without changing the function values at the points $n$ and $n\pm\frac1{n^2}$, the mean value theorem will ensure that for each $n\ge 2$ there will be a point $x\in\left[n-\frac1{n^2},n\right]$ where $f\,'(x)=n$, and therefore $f\,'(x)$ will not approach $0$ as $x\to\infty$.
A: Dan Shved counterexample provide a complete answer, however, if the limit of $\lim_{x\to\infty}f'(x)$ does exist, then $\lim_{x\to\infty}f'(x)$ must be equal to zero. To see this we can apply Lagrange's mean value theorem, we get 
$$\frac{f(x_n+h)-f(x_n)}{h}=f'(c_n)$$ 
where $c_n\in(x_n , x_n+h)$ for any monotone sequence $x_n$, where $\lim_{x\to\infty}x_n=\infty$ and  $h=1$. Since $\lim_{n\to\infty}c_n=\infty$ we get
$$\begin{array}{ll}
\lim_{n\to\infty}f'(c_n) &= \lim_{n\to\infty}\left(f(x_n+1)-f(x_n)\right) \\\\
                         &=0.
\end{array}$$
and this implies that if the limits of $\lim_{x\to\infty}f'(x)$ exists and
$\lim_{x\to\infty}f(x)$ is finite, then $\lim_{x\to\infty}f'(x)=0$. 
