If $f''(x_0)$ exists then $\lim_{x \to x_0} \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$
Prove: if $f''(x_0)$ exists then $\lim\limits_{x \rightarrow x_0} \dfrac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$.
I'm not exactly sure how Taylor's theorem fits into all this, but I found the first derivative, and I haven't been able to progress past that.
We have $$ f(x_0+h)=f(x_0)+f'(x_0)h+\frac{1}{2}f''(x_0)h^2+o(h^2), $$ $$ f(x_0-h)=f(x_0)-f'(x_0)h+\frac{1}{2}f''(x_0)h^2+o(h^2) $$ Adding together we get $$ \frac{f(x_0+h)+f(x_0-h)-2f(x_0)}{h^2}=f''(x_0) + \frac{o(h^2)}{h^2} $$