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?

Always go to Taylor's theorem, which says that if $f(x)$ is differentiable, then
$f(x+h)=f(x)+f'(x)h+o(h)$. 
Similarly if $f$ is $n$ times differentiable, you have:
$$f(x+h)=f(x)+f'(x)h+\cdots+\frac{f^{(n)}(x)}{n!}h^n+o(h^n).$$
Going to the $n=1$ example, notice that even if if is not twice differentiable, the theorem still applies: the error will be $o(h)$. On the other hand, this theorem will fail at any point $x$ where $f$ is not differentiable. You have a function $\phi(x)=\int_0^x\xi^{1/2}\mathrm{d}\xi$. Integrals like this are well known to be differentiable (by fundamental theorem of calculus). But, you seem to be differentiating it twice, and look what happens: $\phi'(x)=x^{1/2}$, 
$$\phi''(x)=\frac{1}{2x^{1/2}}.$$ 
$\phi''(0)$ doesn't exist: the derivative blows up there, hence you're seeing your fraction blow up. 
