Suppose that $f$ has continuous second derivatives. How do I show that
$\frac{f(x+h) + f(x-h) - 2f(x)}{h^2}$
and
2\frac{f(x+h) - f(x) - f'(x)h}{h^2}
both tend to f''(x) as $h \rightarrow 0$?
For the first expression, I can rewrite it as
$\lim_{h \rightarrow 0} \frac{1}{h}(\frac{f(x+h) - f(x)}{h} - \frac{f(x) - f(x-h)}{h})$
which I can sort of see should tend to f''(x), but I can't seem to show it rigorously. For the second expression, I can rewrite it as
\lim_{h\rightarrow 0} \frac{(f(x+h)-f(x))/h - f'(x)}{h}
I'm not sure where the factor of 2 comes in, but I guess it should have to do with the fact that we're trying to take limits "simultaneously" for f' and f''. Can anyone help? Thanks.