Stationary points of a family of functions which converge uniformly. Here's the motivation of my question:

Consider the family of functions $\mathrm{f}_n(x) := \frac{1}{n}\sin(nx)$. It is easy to show that $\mathrm{f}_n$ converges uniformly to the zero function. 

To prove this, we need to show that for all $\varepsilon &gt; 0$, there exists a positive integer $N$ such that $|\mathrm{f}_n(x) - 0| &lt; \varepsilon$ for all $x \in \mathbb{R}$ and $n &gt; N$.

$$\left|\frac{1}{n}\,\sin(nx)-0\right| = \left|\frac{1}{n}\,\sin(nx)\right|
=\left|\frac{1}{n}\right| \cdot \left|\sin(nx)\right| \le \frac{1}{|n|} \cdot 1=\frac{1}{n}$$

Hence $|\mathrm{f}_n(x) - 0| &lt; \varepsilon$ if $\frac{1}{n} &lt; \varepsilon$ and so $|\mathrm{f}_n(x) - 0| &lt; \varepsilon$ for all $n &gt; \frac{1}{\varepsilon}$.

This example is used often to show that derivatives are not inherited from uniformly convergent sequences. For example, $\mathrm{f}_n{'}(0) = 1$ or all positive integers $n$, and yet the zero function has derivative zero everywhere.

Here's my build-up to my question:

I'm interested in when the derivatives of the limiting functions agree with the limit function.
$$\mathrm{f}_n{'}(x) = \cos(nx)$$
The limit function, i.e. the zero function, has zero derivative everywhere. Let $$D_n := \{ x \in \mathbb{R} : \mathrm{f}_n{'}(x) = 0 \}$$
For a fixed $n$, we have $\mathrm{f}_n{'}(x) = 0$ if, and only if, 
$$nx \in \left\{\frac{\pi}{2} + \pi k : k \in \mathbb{Z} \right\}$$
It would seem to me then that
$$D_n = \left\{ \left(\frac{1+2k}{2n}\right)\!\pi : k \in \mathbb{Z} \right\}$$

Here's my question:

What can be said about the set
$$\lim_{n \to \infty} D_n$$

My instinct is that it will be countable and dense in the reals, just like the rationals are. 

EDIT: For example, for any $x \in \mathbb{R}$ and any $\varepsilon &gt; 0$ can we find an $N \in \mathbb{Z}^+$ for which an element of $D_n$ is within $\varepsilon$ of $x$ for all $n &gt; N$?
Counterexample: $f_n(x) = x/n \to 0$ uniformly on $[0,1],$ but $f_n(x)$ has no stationary points.
Counterexample: $f_n(x) = x/n \to 0$ uniformly on $[0,1],$ but $f_n(x)$ has no stationary points.