Limit formula for the second derivative Suppose that $f$ is a real-valued function which is twice differentiable in the interval $(-1,1)$. Does the following hold?:
$$\lim_{h \to 0} \frac{f(h) - 2f(0) + f(-h)}{h^2} = f''(0)$$ 
If $f''(x)$ is continuous this follows from Taylor's theorem with the
Lagrange form of the remainder.  What if $f''(x)$ is not continuous?
-- Thanks.
 A: Just so an answer is put here rather than in the comments:
As noted in the original question, the desired limit formula is a well-known, or at least straightforward, consequence of Taylor's theorem with Lagrange remainder in cases where $f\in C^2$, i.e. if we have some continuity of the 2nd derivative in a neighbourhood of $0$.
As pointed out by Nate Eldredge in the comments: for functions which are merely twice differentiable at the origin, i.e. where the second derivative exists at $0$ but might not be continuous in a neighbourhood of $0$, one still has Taylor's theorem (of order $2$) with Peano remainder:
$$
f(x) = f(0) + f'(0)x + \frac{1}{2}f''(0)x^2 + h(x)x^2
$$
where $h(x)\to 0$ as $x\to 0$. This is enough to obtain the desired limit formula.
Remark. I took the liberty of contacting the OP and he said:

I am teaching the first real analysis course
  for math majors.  In class I
  proved that if you know the value of a twice
  differentiable function at three points you
  know the value of the second derivative at
  some point ... and I assumed you need continuity
  of the second derivative to prove the
  limit formula.

