If a pdf $f$ is differentiable, does it have to satisfy $\lim _{x\to \infty}f'(x)\to0$?
At first I thought this was obvuously true, but I think it can be false if the CDF has infinitely many inflection points (if that's even possible - see picture).
That is, just because the integral (CDF) tends to $1$ doesn't mean that it eventually coincides with the horizoal line $y=1$.
On
Let $T$ be the "tent function" defined by
$$T_n(x)=max(0,n-n^4*|x|)$$
Its support is $[-\dfrac{1}{n^3},\dfrac{1}{n^3}]$.
As its maximum is at height $n$, the area under this curve is clearly $\dfrac{1}{n^2}$
If we define $f$ by $f(x)=\dfrac{6}{\pi^2}\sum_{n=0}^{\infty}T_n(x-n)$ we have a pdf with arbitrary large slopes as $n \rightarrow \infty.$
Regarding the title: it's possible for a CDF to be strictly monotone over the whole line, for example the CDF of a normal distribution has this property.
Regarding the question in the body: it's also possible for a pdf to fail to go to zero, and for its derivative to fail to go to zero. The classic example of this is something like $f(x)=\sum_{n=1}^\infty 2^{-n} g(x-n)$ where $g(x)=\begin{cases} 0 & x<0 \\ 4x & 0 \leq x \leq 1/2 \\ 4-4x & 1/2< x \leq 1 \\ 0 & x>1 \end{cases}$.
(The graph of $g$ is just a triangle with area $1$.) This example is not differentiable everywhere, but it can be smoothed as needed while still keeping it a pdf. For instance a $C^1$ version is $g(x)=\begin{cases} 0 & x<0 \\ \frac{1}{C} x^2(x-1)^2 & 0 \leq x \leq 1 \\ 0 & x>1 \end{cases}$ where $C=\int_0^1 x^2(x-1)^2 dx$.