Wikipedia wrong? Convergence of finite difference Update: I have edited the Wikipedia page, so that the mistake no longer appears.
On the Wikipedia article for "Finite difference" there is the claim 

Assuming that $f$ is continuously differentiable, [we have] $$
 \frac{f(x+h) - f(x)}{h} - f'(x) = O(h) \quad \text{as}\,\, h \to
 0.\tag{1}  $$ The central difference gives a more accurate approximation. [Supposing that $f$ is $C^2$] $$ \frac{f\left(x+\frac12
 h \right)- f \left( x - \frac12 h \right)}{h} - f'(x) = O(h^2). \tag{2} $$

I think these are false. In the case of (1), consider $\phi(x) = \int_0^x \xi^{1/2} d\xi$. It is $C^1$ on $[0,\infty)$, but 
$$
\frac{\frac{\phi(0+h) - \phi(0)}{h} - \phi'(0)}{h} \quad \text{is unbounded as } h \to 0.
$$
I believe that (2) is false for similar reasons, and I think a counterexample is $\psi(x) = \int_0^x \int_0^\xi \eta^{1/2}d\eta d\xi$, but I haven't worked it out.
Could someone tell me what the correct statement of these things is, or explain to me why I'm wrong and Wikipedia is right?

You are right. In the first case, for
$$\frac{f(x+h)-f(x)}{h} - f'(x)$$
you only have an $o(1)$ bound. A function like $f(x) = x\cdot\lvert x\rvert^\alpha$ for $0 < \alpha < 1$ is continuously differentiable on all of $\mathbb{R}$, but at $0$ the difference quotient converges only of the order $\lvert h\rvert^{\alpha}$ to the derivative.
In the second case, choosing $1 < \alpha < 2$ gives a twice continuously differentiable function with
$$\frac{f\left(x + \tfrac{h}{2}\right) - f\left(x - \tfrac{h}{2}\right)}{h} - f'(x) \in \Theta(h^{\alpha}).$$
The order of convergence under the assumption of differentiability resp. twice differentiability is
$$\frac{f(x+h)-f(x)}{h} - f'(x) \in o(1)$$
resp.
$$\frac{f\left(x + \tfrac{h}{2}\right) - f\left(x - \tfrac{h}{2}\right)}{h} - f'(x) \in o(h),$$
nothing better is to be had without stronger assumptions.
