Analysis Question from Berkeley Problems in Mathematics I'm wondering if the following is correct. The original question(1.1.21, Fa96) asks to prove that
\begin{align*}
f''(x) = \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2}
\end{align*}
I thought it was straightforward and proposed the following solution:
\begin{align}
f'(x) &= \lim_{h\rightarrow0}\dfrac{f(x+h) - f(x)}{h} 
\\
f'(x-h)&= \lim_{h\rightarrow0}\dfrac{f(x) - f(x-h)}{h}\\
f''(x) &=\lim_{h\rightarrow0} \dfrac{f'(x) - f'(x-h)}{h} \\
\end{align}
Substitute $f'(x)$ and $f'(x-h)$ and we'll end up with our desired identity. This solution is different from the one proposed in the book so I want to make sure it is correct. 
 A: Strictly speaking no your solution is not correct.
We have a function
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
but note if we try to evaluate $f'(x-h)$ then the $h$ inputted and the $h$ in the limit are two different variables (one is being taken to $0$ while the other is being taken to be constant). So really what we have is
$$
f'(x-h') = \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h}
$$
and now looking at $f''$ we get
$$
f''(x) = \lim_{h' \to 0} \frac{f'(x) - f(x-h')}{h'} = \lim_{h' \to 0} \frac{\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h}}{h'}
$$
And now assuming sufficient differentiability of $f$ we get
$$
f''(x) = \lim_{h' \to 0} \lim_{h \to 0} \frac{f(x+h)-f(x)-f(x-h'+h)+f(x-h')}{h h'}
$$
but this requires justification in that you can interchange these limits (or take them both to $0$ at the same rate).
A: Recently came across this in wikipedia : http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
We need to prove that : 
\begin{align*}
f''(x) = \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2}
\end{align*}
Now, observe that this limit is of the ${0\over 0}$ form.  So, we can apply the l'hopitals rule. So, we differentiate with respect to $h$ to get
\begin{align*}
\lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2}= \lim_{h\rightarrow0}\dfrac{f'(x+h) - f'(x-h) }{2h}=f''(x)\end{align*}
A: Hint: Another approach is to use Taylor series as

$$ f(x+h) \approx f(x)+f'(x)h+\frac{f''(x)}{2!}h^2 $$
$$ f(x-h) \approx f(x)-f'(x)h+\frac{f''(x)}{2!}h^2. $$

See a related problem.
