Evaluation Of Iterated Limit Let $f: \textrm{dom}(f) \rightarrow \mathbb{R}$.
Let $a \in \textrm{dom}(f)$.
Assume $f''(a)$ exists.
Want To Prove $f''(a) = \displaystyle\lim_{h \rightarrow 0} \frac{f(a + h) + f(a - h) - 2f(a)}{h^2}$ .
$f''(a)$
$= \displaystyle\lim_{h \rightarrow 0} \frac{f'(a + h) - f'(a)}{h}$
$= \displaystyle\lim_{h \rightarrow 0} \frac{\displaystyle\lim_{k \rightarrow 0} \frac{f(a + h + k) - f(a + h)}{k} - \lim_{k \rightarrow 0} \frac{f(a + k) - f(a)}{k}}{h}$
$= \displaystyle\lim_{h \rightarrow 0} \frac{\displaystyle\lim_{k \rightarrow 0} \bigg( \frac{f(a + h + k) - f(a + h)}{k} - \frac{f(a + k) - f(a)}{k} \bigg)}{h}$
$= \displaystyle\lim_{h \rightarrow 0} \frac{\displaystyle\lim_{k \rightarrow 0} \frac{f(a + h + k) - f(a + h) - f(a + k) + f(a)}{k}}{h}$
$= \displaystyle\lim_{h \rightarrow 0} \Bigg( \! \lim_{k \rightarrow 0} \frac{f(a + h + k) - f(a + h) - f(a + k) + f(a)}{hk} \Bigg)$
In order to get the desired result, I would like to replace $k$ with $-h$. I did this because as $h \rightarrow 0$ and $k \rightarrow 0$, $k \rightarrow -h$.
However, this reasoning of "$h \rightarrow 0$ and $k \rightarrow 0$" would be more applicable in the case of a two-variable limit as $(h, k) \rightarrow (0, 0)$ rather than an iterated limit as seen in the expression, which may not be equal: Link .
So how can I justify replacing $k$ with $-h$?
I thought about using the Moore-Osgood Theorem, which says the iterated limit is equal to the two-variable limit under a sufficient condition. However, I am unfamiliar with multivariable calculus and I am not sure how I would use it in this case. Unfortunately, I also cannot find a proof that follows Wikipedia's description: Link .

According to Taylor’s theorem it exists a map $r_2$ defined in a neighborhood of zero such that

*

*$f(a+h)=f(a)+f^{\prime}(a)h +\frac{1}{2}f^{\prime\prime}(a)h^2+h^2r_2(h)$

*$\lim\limits_{h \to 0} r_2(h)=0$
You then get the desired conclusion from the equality
$$\frac{f(a + h) + f(a - h) - 2f(a)}{h^2}=f^{\prime\prime}(a)+r_2(h)+r_2(-h)$$
