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 $\varepsilon > 0$, there exists a positive integer $N$ such that $|\mathrm{f}_n(x) - 0| < \varepsilon$ for all $x \in \mathbb{R}$ and $n > N$.
$$\left|\frac{1}{n}\,\sin(nx)-0\right| = \left|\frac{1}{n}\,\sin(nx)\right| =\left|\frac{1}{n}\right| \cdot \left|\sin(nx)\right| \le \frac{1}{|n|} \cdot 1=\frac{1}{n}$$
Hence $|\mathrm{f}_n(x) - 0| < \varepsilon$ if $\frac{1}{n} < \varepsilon$ and so $|\mathrm{f}_n(x) - 0| < \varepsilon$ for all $n > \frac{1}{\varepsilon}$.
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 $\varepsilon > 0$ can we find an $N \in \mathbb{Z}^+$ for which an element of $D_n$ is within $\varepsilon$ of $x$ for all $n > N$?
Counterexample: $f_n(x) = x/n \to 0$ uniformly on $[0,1],$ but $f_n(x)$ has no stationary points.