A formula for second derivative I want to prove that if $f''$ is continuous at $x_0$, then $$f''(x_0)=\displaystyle\lim_{h\to 0}\dfrac{f(x_0+2h)+f(x_0)-2f(x_0+h)}{h^2}$$
Any hint to prove it?
I can't use l'Hopital or Taylor.
Thanks.

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