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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.__