I remember reading somewhere that for a $f\in C_c^1(\mathbb{R})$, by definining
$$f_n(x) := \frac{f(x+1/n)-f(x)}{1/n},$$
we have that $f_n(x) \longrightarrow f'(x)$ uniformly.
I'm pretty sure the compact support condition needs to hold for this, but when I tried proving this claim, I can't see where I assumed this, or where the proof fails:
For a given $\epsilon > 0$, choose $N$ so that $|x-y| < 1/N$ implies $|f'(x)-f'(y)| < \epsilon$. Then for any $x_0\in \mathbb{R}, n>N$, from the MVT we have a $\gamma_{x_0,n} \in (x_0, x_0 + 1/n)$ such that $|f_n(x_0)-f'(x_0)| = |f'(\gamma) - f'(x_0)|$. But since $|\gamma - x_0| < 1/n < 1/N$, it follows that $|f_n(x_0)-f'(x_0)| < \epsilon$.
So which step in the proof is incorrect/uses compact support-ness?