Here's Exercise 11 in Baby Rudin:
Suppose $f$ is defined in a neighborhood of $x$, and suppose $f^{\prime\prime}(x)$ exists. Show that \begin{equation}\label{11.0} \lim_{h \to 0} \frac{f(x+h)+ f(x-h)-2f(x)}{h^2} = f^{\prime\prime}(x) \end{equation} Show by an example that the limit may exist even if $f^{\prime\prime}(x)$ does not.
I had no trouble proving the statement but I am having trouble coming up with an example. Initially, I thought of:
$f(x) = \begin{cases} x+1 & \text{if $x<0$} \\ 0 & \text{if $x=0$} \\ x-1& \text{if $x>0$} \end{cases} $
Then, as $h \to 0$, $\lim_{h \to 0} \frac{f(x+h)+ f(x-h)-2f(x)}{h^2} = \begin{cases} \lim\limits_{h \to 0} \frac{(x+h+1)+ (x-h+1)-2x-2}{h^2} = \lim\limits_{h \to 0} \frac{0}{h^2}=0 & \text{if $x<0$} \\ \lim\limits_{h \to 0} \frac{0}{h^2}=0 & \text{if $x=0$} \\ \lim\limits_{h \to 0} \frac{(x+h-1)+ (x-h-1)-2x+2}{h^2} = \lim\limits_{h \to 0} \frac{0}{h^2}=0& \text{if $x>0$} \end{cases}$
Note that $f(0-) = f(0+) = \lim\limits_{h \to 0} f(0) = 0$. However, this also leads me to:
$f''(x) = \begin{cases} 0 & \text{if $x<0$} \\ 0 & \text{if $x=0$} \\ 0& \text{if $x>0$} \end{cases}$
Is my example even correct at all? If not, can someone suggest a better example and show how $f''(x)$ does not exist (presumably at some point)?
how about $f(x) = \begin{cases} -x^2 & \text{if $x<0$} \\ x^2 & \text{if $x\geq 0$} \\ \end{cases} $.
second derivative at $0$ does not exist but the limit zero
$$\lim_{h \to 0} \frac{f(x+h)+ f(x-h)-2f(x)}{h^2}$$ at $x=0$ exists (and is equal to $0$).