differentiation proof question

f(x) is differentiable twice at x_0

prove that:

[math]

f''(x_0 ) = \mathop {\lim }\limits_{h \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}

[/math]

i tried to solve it but i cant get to the asked expression

[math]

g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}\\

[/math]

[math]

g''(x_0) = \lim_{h \to 0} \frac{g'(x_0 + h) - g'(x_0)}{h}\\

[/math]

[math]

g''(x_0) = \lim_{h \to 0} \frac{\frac{g(x_0 + 2h) - g(x_0+h)}{h} - \frac{g(x_0 + h) - g(x_0)}{h}}{h}\\

[/math]

[math]

g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - g(x_0+h)} - {g(x_0 + h) - g(x_0)}}{h^2}\\

[/math]

[math]

g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - 2g(x_0+h)} { - g(x_0)}}{h^2}\\

[/math]

??

Edited February 9, 200916 yr by transgalactic