Expert: Alon Mandes - 11/8/2009

Question
if f'' is continuous show that

Lim
h->0

f(x + h) -2f(x) + f(x-h)
------------------------
h^2

= f''(x)


I keep attempting to prove this using l'hospitals rule but I keep getting 0/1 as my answer. Thanks for your time.


Answer
The definition of f'(x) is :
Lim{h->0} [f(x+h)-f(x)]/h .
Therefore,
The definition of f''(x) will be :
Lim{h->0} [f'(x+h)-f'(x)]/h . Which is :
f''(x)=Lim{h->0} (   [Lim{h->0}[f(x+2h)-f(x+h)]/h   -   Lim{h->0}[f(x+h)-f(x)]/h    )/h² . Or


                        f(x+2h)-f(x+h)                  f(x+h)-f(x)
             Lim{h->0}  -------------    -   Lim{h->0} --------------
                              h                              h
Lim{h->0} -------------------------------------------------------------   =
                                         h

                                        
              f(x+2h)-f(x+h)-f(x+h)+f(x)
Lim{h->0} ---------------------------------- =
                        h²

             f(x+2h)-2f(x+h)+f(x)
Lim{h->0} ------------------------- .
                 h²

Note that f'(x) can also be written as : f'(x)=Lim{h->0} [f(x+h)-f(x-h)]/2h .

Now let's chick out the form we are interested in :

         f(x+h)-2f(x)+f(x-h)
Lim{h->0} ------------------- . This form satisfy 0/0 thus we may use  L'hospital's rule :
                 h²


             f'(x+h)-2f'(x)+f'(x-h)
= Lim{h->0} ------------------------- . If we substitute the definition of f' we get :
                     2h


             f(x+2h)-f(x+h)-2[f(x+h)-f(x)]+f(x)-f(x-h)
= Lim{h->0} --------------------------------------------- =   
                     2h²

             f(x+2h)-f(x+h)-2f(x+h)+2f(x)+f(x)-f(x-h)
= Lim{h->0} --------------------------------------------- =   
                     2h²


             f(x+2h)-2f(x+h)+f(x) + 2f(x)-f(x+h)-f(x-h)
= Lim{h->0} --------------------------------------------- =   
                     2h²

            f(x+2h)-2f(x+h)+f(x)      2f(x)-f(x+h)-f(x-h)
= Lim{h->0} --------------------- + ------------------------ =   
                     2h²                   2h²

= f''(x)+0=f''(x) .

Alon.