Cauchy's Mean Value Theorem and Higher Order Derivatives My teacher told me that one can derive the formula for second order derivative limit form using CMVT and this is how I tried to construct it, could someone please tell me if I did it correctly or not? Proving $\lim_{h\to0} \frac{f(a+h)+f(a-h)-2f(a)}{h^2}$ $=$ $f''(a)$.
My attempt
Let $f,g$ be defined on a $[0, h]$, let $F(x)= f(a+x)+f(a-x)-2f(a)$ and $G(x) = x^2$. By CMVT, there exists $k$ in $[0, h]$ such that $\frac{F(h)}{G(h)}$ = $\frac{f'(a+k)-f'(a-k)}{2k}$. Then, $\lim_{h\to0}\frac{F(h)}{G(h)}$ = $\lim_{k\to0} \frac{f'(a+k)-f'(a-k)}{2k}$ = $\lim_{k\to0} \frac{f''(a+k)+f''(a-k)}{2}$, where the second equality comes from L'Hopital's Rule. Then, $\lim_{k\to0} \frac{f''(a+k)+f''(a-k)}{2}$ = $f''(a)$. So the equality is proven.
My question
Am I allowed to take limit as $h$ tend to $0$ on one side and the other side as $k$ tend to $0$?
 A: Idea is correct. 
However, the functions as defined are inconsistent with the variables (at least when posted) and using L'Hospital both beats the purpose and assumes more than may actually be given (does the second derivative exist at points other than $a$? Is it continuous at $a$?). 
I'll first write down a complete statement, then repeat the proof with the necessary changes. 

Assume $f$ is differentiable in some open interval containing $a$ and have second derivative at $a$. Then 
$$ \lim_{h\to0} \frac{f(a+h)+f(a-h)-2f(a)}{h^2}=f''(a).$$

Proof.  FIX $a$. 
Let $F(h) = f(a+h) +f(a-h) - 2f(a)$ and $G(h) = h^2$. 
Note that $F'(h) = f'(a+h) - f'(a-h),~G'(h)=2h$. Then 
$$\frac{F(h)}{G(h)}=\frac{F(h) - F(0)}{G(h)-G(0)} \overset{_{CMVT}}{=} \frac{F'(c_1)}{G'(c_1)}=(*)$$
Observe that
$$(*) =  \frac{f'(a+c_1) - f'(a-c_1)}{2c_1} =\frac 12\left ( \frac{f'(a+c_1)-f(0)}{c_1} + \frac{f'(a-c_1)-f(0)}{-c_1}\right).$$ 
Now  as $h\to 0$, $c_1\to0$, and therefore it follows from the definition of the derivative of a function at a point, in this case, the function $f'$ at $a$,  that 
the limit of $(*)$ as $h\to0$ exists and is equal to $f''(a)$.
