Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$ Let $f''$ be continuous on $\mathbb{R}$. Show that 
$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$


My workings 
$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\lim_{h\to0}\frac{f(x+h)-f(x)-[f(x)-f(x-h)]}{h^2}=\frac{\lim_{h\to0}\frac{f(x+h)-f(x)}{h}-\lim_{h\to0}\frac{f(x)-f(x-h)}{h}}{\lim_{h\to0}h}$$
By the definition of derivative, I move on to the next step. Also, I observe that everything in this question as continuous and differentiable up to $f''(x)$.
$$=\frac{f'(x)-f'(x-h)}{\lim_{h\to0}h}$$
I do not know how to justify the next move but, 
$$=\lim_{h\to0}\frac{f'(x)-f'(x-h)}{h}$$
Then by the definition of derivative again, 
$$=f''(x-h)$$
Which is so close to the answer. So I shall assume that since $h\to0$ for $x-h$, therefore $x-h=x$? And so, 
$$=f''(x)$$


I think i made a crapload of generalization and fallactic errors... I also have another way, which was to work from $f''(x)$ to the LHS. But I realised I assume that the h were the same for $f'(x)$ and $f''(x)$.
Is it normal to be unable to solve this question at the first try? Or am I just too weak in mathematics?

Apply Taylor's formula in the form
$f(x+h) = f(x) + h f'(x) + h^2 f''(x)/2 + o(h^2 f''(x))$.
