Prove that $\lim_{h\to 0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2} = g''(a)$

Question

How do I prove that

$$\displaystyle\lim_{h\to0} \dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} = g''(a),$$

where $g$ is of class $C^2$?

My attempt:

By Mean Value Theorem, there are $c_0\in (a-h, a)$ and $c_1\in(a, a+h)$ such that

$$g(a)-g(a-h) = g'(c_0)\cdot h$$

$$g(a+h)-g(a) = g'(c_1)\cdot h$$

Then

$\displaystyle\lim_{h\to0} \dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} = \displaystyle\lim_{h\to0} \dfrac{g(a+h)-g(a)-(g(a)-g(a-h))}{h^2} = \\ \displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{h}= \displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{c_1-c_0}\cdot\dfrac{c_1-c_0}{h}=\\ \displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{c_1-c_0}\cdot\displaystyle\lim_{h\to0}\dfrac{c_1-c_0} {h} $

The first limit is equals to $g''(a)$. Then the second limit should be equals to $1$, but I can't see why.

Answer 1

If we know $g''(x)$ exists for an open interval $I$ containing $a$, $$\lim_{h\to0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to0} \frac{g'(a+h)-g'(a-h)}{2h} =\displaystyle\lim_{h\to0} \frac{g'(a+h)-g'(a)+g'(a)-g'(a-h)}{2h} =g''(a) $$, by L'Hopital Rule

Answer 2

By Taylor's expansion

• $g(a+h)=g(a)+g'(a)h+\frac12g''(a)h^2+o(h^2)$
• $g(a-h)=g(a)-g'(a)h+\frac12g''(a)h^2+o(h^2)$

and adding up

$$g''(a)=\frac{g(a+h)-2g(a)+g(a-h)}{h^2}+\frac{o(h^2)}{h^2}$$

then take the limit.

Answer 3

\begin{align*} g(a+h)&=g(a)+g'(a)h+\dfrac{1}{2}g''(\xi_{h})h^{2}\\ g(a-h)&=g(a)-g'(a)h+\dfrac{1}{2}g''(\eta_{h})h^{2}, \end{align*} where $\xi_{h}$ lies in between $a$ and $a+h$, and $\eta_{h}$ lies in between $a$ and $a-h$, so \begin{align*} \dfrac{1}{h^{2}}[g(a+h)+g(x-h)-2g(a)]=\dfrac{1}{2}(g''(\xi_{h})+g''(\eta_{h}))\rightarrow\dfrac{1}{2}(g''(a)+g''(a))=g''(a) \end{align*} by the continuity of $g''$.

Answer 4

Note that $$\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} =\dfrac{\dfrac{g(a+h)-g(a)}{h} -\dfrac{g(a)-g(a-h)}{h}}{h},$$ so that quotient is basically the slope of $g'$ at $x=a$ for $h$ very small.