differantiation proof question exlanation problem..

[transgalactic]

if f(x) is differentiable on x_0
prove that
http://img102.imageshack.us/img102/3189/51290270sj1.th.gif (http://img102.imageshack.us/my.php?image=51290270sj1.gif)

??

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

now let u = x+h \Rightarrow x = u-h

So f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h}

Since h \to 0, f'(u-h) \to f'(u)

Therefore f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}

f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)

2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h}

Hence, f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}


i understand this solution but
how to pick those variables??
and why it didnt use the point x_0
that they presented

[transgalactic]

there is another definition to limit that uses x_0 that i was given

f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}


is there a way to use this definition like you did before
what variables to pick so it will go on the "h" defnition??

[transgalactic]

????