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?
 A: Try applying L'Hospital's Rule to $h$, that is, differentiate with respect to $h$.
A: When you use that $f'(x) = \lim_{h \to 0} \frac {f(x+h)-f(x-h)}{2h}$, you get that $f''(x) = \lim_{k \to 0} \lim_{h \to 0} \frac {f(x+h+k)-f(x+h-k)-f(x-h+k)+f(x-h-k)}{4hk}$.
Let $g_x(h,k)$ be that expression : $f''(x) = \lim_{k \to 0} \lim_{h \to 0} g_x(h,k)$, while what you are given is $\lim_{h \to 0} g_x(h/2,h/2)$.
Both limits are going to $(0,0)$ but not along the same path. So it makes sense that they should be equal under certain conditions, namely if the function $g_x$ can be continuously extended at the point $(h=0,k=0)$ and on the two axis $h=0$, $k=0$: then no matter what path you take in your limit to $(0,0)$, you will get the same result.
In order to show that it is the case, you need to use the fact that $f''$ is continuous. Apply the mean value theorem to the functions $g_{x,h} : k \mapsto f(x+h+k) - f(x-h+k)$ :
forall $h,k$, there is a $k'$ such that $|k'|\le |k|$ and $g_{x,h}(k) - g_{x,h}(-k) = 2kg'_{x,h}(k')$, which means that $g_x(h,k)$ simplifies to $\frac{g'_{x,h}(k')}{2h} = \frac{f'(x+h+k') - f'(x-h+k')}{2h}$.
Note that with the continuity of $f'$, this implies that for $h \neq 0$, $g_x$ can be continuously extended at $g_x(h,0)$ by $g_x(h,0) = \frac {f'(x+h)-f'(x-h)}{2h}$ (and similarly on the other axis)
Next we can apply the mean value theorem again, to all the functions $h \mapsto f'(x+h+k')$:
forall $h,k$ there are some $h',k'$ such that $|h'| \le |h|, |k'| \le |k|$, and $g_x(h,k) = f''(x+h'+k')$.
Then, we use the continuity of $f''$ to conclude that $\lim_{(h,k) \to (0,0)} g_x(h,k) = \lim_{(h,k) \to (0,0)} f''(x+h'+k') = f''(x)$
A: Apply Taylor's formula in the form
$f(x+h) = f(x) + h f'(x) + h^2 f''(x)/2 + o(h^2 f''(x))$.
A: Your problem hinges on L'Hospital's Rule for $h$ and the following formula for derivative which known as the symmetric formula
$$ f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}  \,,$$
which implies
$$ f''(x) = \lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}\,. $$
See here for details of deriving the above formula.
A: Apply Cauchy mean value theorem for $p(h)=f(x+h)+f(x-h)-2f(x)$, and $g(h)=h^2$, in $[0,h]$, twice. 
Since $p(0)=p'(0)=g(0)=g'(0)=0$, you can get
$\frac{p(h)}{h^2}=\frac{p''(z)}{2}=\frac{f''(x+z)+f''(x-z)}{2}\ \text{for z is strictly beteween h and 0}$
Therefore, if you take h goes to zero, then z also goes to zero. Finally, use the contiunuity condition of $f''$.
