Suppose f is defined in a neighborhood of x, and suppose f '' (x) exists. Show that: lim [f(x+h)+f(x-h)-2f(x)] / h^2 = f''(x). h->0 Show by an example that that the limit may exist even if f '' (x) may not. (hint: use lHopital's Theorem). Proof: f '' (x) exists implies f ' (x) exists and is differentiable which implies f(x) exists and is differentiable. Thus, f '' (x) = lim [f ' (x) - f ' (t)] / (x-t) (def of derivative) x->t Doing a substitution h = x - t lim [f ' (x+h) - f ' (x)] / h (def of derivative) h->0 = lim [f(x+h) - f(x) - (f(x) - f(x-h)] / h / h h->0 = lim [f(x+h) + f(x-h) - 2f(x)] / h^2 (*) h->0 QED What I'm having trouble with is the example of where the limit exists but f '' (x) does not. Either using the original limit/derivative definition or the end of the proof (*) I get 0/0 or another indeterminate form.