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