Proof identity of differential equation I would appreciate if somebody could help me with the following problem:
Q: $f''(x)$ continuous in $\mathbb{R}$ show that 
$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$ 
 A: Use  L'Hospital's rule
$$ \lim_{h\to 0}\frac{F(h)}{G(h)}=\lim_{h\to 0}\frac{F'(h)}{G'(h)}$$
You can use the rule if you have $\frac{0}{0}$ result.
You need to apply it 2 times
$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}= \lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}=\lim_{h\to 0}\frac{f''(x+h)+f''(x-h)}{2}=f''(x)$$ 
A: Hint: what is the limit definition of a derivative? Try writing it out, first applying it to $f(x)$, and then applying it to $f'(x)$. Can you apply this definition several times to get what you are looking for? (Note that you must be careful in this argument, as it requires the continuity of $f''(x)$!)
A: Let $N_{\delta}(c)=(c-\delta,c+\delta)$ be a neighbourhood of $c$ where$f''$ is continuous.Then for any $0<h<\delta$  we have from Taylor's Theorem with $R_2$ , the remainder in Lagrange's form   
$f(c+h)=f(c)+hf'(c)+\frac{h^2}{2!}f''(c+\theta h)$  , ($0<\theta<1$  )
$f(c-h)=f(c)-hf'(c)+\frac{h^2}{2!}f''(c-\theta 'h)$ , ($0<\theta '<1$  )   
and $\displaystyle\lim_{h\to 0}f''(c+\theta h)=f''(c)$ and $\displaystyle\lim_{h\to 0}f''(c-\theta ' h)=f''(c)$.  
Using these results you can easily carry out the calculation I think.
