Limit identity for $f''(x)$ proof

Question

Problem statement:

Let $f: (a,b) \rightarrow \mathbb{R}$ be given. If $f''(x)$ exists, prove that $$\lim_{h \rightarrow 0} \frac{f(x-h)-2f(x)+f(x+h)}{h^2} = f''(x).$$

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Is the following reasoning correct?

Consider the following facts:

\begin{align} \lim_{h \rightarrow 0} f(x + h) &= \lim_{h \rightarrow 0} f(x + 2h) \\ \lim_{h \rightarrow 0} f(x + h) &= f(x) = \lim_{h \rightarrow 0} f(x) \\ \lim_{h \rightarrow 0} f(x - h) &= f(x) = \lim_{h \rightarrow 0} f(x) \\ \end{align}

It follows as such: \begin{align*} \lim_{h \rightarrow 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} &= \frac{\lim_{h \rightarrow 0}f(x+h) - 2\lim_{h \rightarrow 0}f(x) + \lim_{h \rightarrow 0}f(x-h)}{\lim_{h \rightarrow 0}h^2} \\ &= \frac{\lim_{h \rightarrow 0}f(x+2h) - 2\lim_{h \rightarrow 0}f(x+h) + \lim_{h \rightarrow 0}f(x)}{\lim_{h \rightarrow 0}h^2} \\ &= \lim_{h \rightarrow 0} \frac{f(x+2h) - 2f(x+h) + f(x)}{h^2} \\ &= \lim_{h \rightarrow 0} \frac{\frac{f(x+2h) - 2f(x+h) + f(x)}{h}}{h} \\ &= \lim_{h \rightarrow 0} \frac{\frac{f(x+2h) - f(x+h)}{h} - \frac{f(x+h) - f(x)}{h}}{h} \\ &= \lim_{h \rightarrow 0} \frac{\lim_{h \rightarrow 0}\frac{f(x+2h) - f(x+h)}{h} - \lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h}}{h} \\ &= \lim_{h \rightarrow 0} \frac{f'(x+h) - f'(x)}{h} \\ &= f''(x) \end{align*}

Answer 1

By Taylor-Young formula,

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+h^2\epsilon_1(x)$$

$$f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)+h^2\epsilon_2(x)$$

sum and finish.

Answer 2

Yes, and furthermore you can skip the first two steps by performing substitution \begin{align*} u = x + h \end{align*}

Answer 3

Your argument fails here: $$ \lim_{h \rightarrow 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = \frac{\lim_{h \rightarrow 0}f(x+h) - 2\lim_{h \rightarrow 0}f(x) + \lim_{h \rightarrow 0}f(x-h)}{\lim_{h \rightarrow 0}h^2} $$ that is at the first step.

The right hand side is $0/0$, which is undefined.

The arguments you're using later are similarly wrong. Consider $$ \lim_{x\to0}\frac{x-\sin x}{x^3}= \lim_{x\to0}\frac{\lim_{x\to0}(x-\sin x)}{x^3}=\lim_{x\to0}\frac{0}{x^3}=0 $$ Sorry, the limit is $1/6$.