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**mathematicalbagpiper** Show that $\displaystyle f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

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

But also that:

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

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

So when I substitute these in, I'm instead getting the following:

$\displaystyle \displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$

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

Which I know is just off by an $\displaystyle h$. It seems to me that they get the second derivative formula like this:

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

But I don't really understand why it's that way. That seems like $\displaystyle f''(x-h)$ to me. Why is it done that way?