Limit of a function and its derivative Do there exist functions $f$, where $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} \frac{df(x)}{dx} \neq 0$?
 A: Take $f(x)= \frac{1}{x} \sin(x^2)$.
A: If exists $\lim\limits_{x\to+\infty}{\dfrac{df(x)}{dx}}=a\ne{0}$ then $\forall \varepsilon>0\;\;\exists x_\varepsilon\colon \;\; \forall {x>x_\varepsilon}$
$$a-\varepsilon<{\dfrac{df(x)}{dx}}<a+\varepsilon, \tag{*}$$
which contradicts the mean value (Lagrange's) theorem.
Therefore if $\lim\limits_{x\to+\infty}{\dfrac{df(x)}{dx}}$ exists, it equals $0.$
Edited (more than 8 years later)
At first it should be said that the claim about contradiction with MVT in the previous answer was formulated insufficiently correctly.
More precizely, it means that the existence of nonzero limit $\lim\limits_{x\to+\infty}{f'(x)}$ using the Mean Value Theorem contradicts to the fact that the limit $\lim\limits_{x \rightarrow +\infty} f(x) = 0.$
Note that $({*})$ holds, in particular, for $\varepsilon < \frac{|a|}{2}.$
Now we take an arbitrary $x_1 > x_{\varepsilon}$ and set $x_2 = 2x_{1}.$ Then by the Mean Value Theorem there exists some $\xi, \  x_1 < \xi < x_2,$ such that
$$f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1) =  f'(\xi) \cdot x_1. $$Now it follows that $$\left\vert f(x_2) - f(x_1) \right\vert =  \left\vert f'(\xi) \right\vert \cdot x_1 > x_1 \cdot \min\{ |a-\varepsilon|, \   |a+\varepsilon|\}  $$ which is unbounded for sufficiently large $x_1,$ hence $\lim\limits_{x \rightarrow +\infty} f(x) $ cannot be $0.$
