Definition of Limit of a Function

Question

I'm a little confused by the definition of the limit of a function-on one hand I feel the definition suggests that your limiting variable is shrinking into a little delta ball- on the other hand when it comes to derivatives, of things like $x^2$ for $x \ge 0$ $-x^2$ for $x \le 0$ to show that it isn't second differentiable we consider$f(-h)$ and $f(h)$, where these h's somehow never take on a negative sign. Which is it, and what resolves my confusion?

Answer 1

The values of $h$ take negative and positive signs. The problem is this: The function $$f(x)=\begin{cases}x^2&;x\geq0\\-x^2&;x\leq0\end{cases}$$ has a first derivative of $f'(x) = |x|$.

The reason that $f$ does not have a second derivative is that the functin $g(x)=|x|$ does not have a derivative at $0$, so let's just look at that.

$g$ has a derivative at $x=0$ if the limit $$\lim_{h\to0}\frac{g(0+h)-g(0)}{h} = \lim_{h\to0}\frac{|h|}{h}$$ exists, and here is where the problem is. The thing is that if $h>0$, then $\frac{|h|}{h} = 1$, and for $h<0$, $\frac{|h|}{h}=-1$.

Strictly speaking, for the limit to exist and be equal to $L$, this statement should be true:

$$\forall \epsilon>0\exists \delta > 0\forall h, |h|<\delta: \left|\frac{|h|}{h}-L\right|.$$

However, this statement is not true, because for every $\delta>0$, we can find values $h_1=\frac{\delta}{2}$ and $h_2=-\frac{\delta}{2}$ such that $\frac{|h_1|}{h_1}=1$ and $\frac{|h_2|}{h_2}=-1$.

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I see you are asking about the formula for the second derivative which is $$\frac{f(x+h)+2f(x)+f(x-h)}{h^2}.$$ However, this is not the definition of the second derivative. This is only a formula to calculate the derivative if the derivative exists. Therefore, it is useless in your case and tells you nothing. This link in wikipedia tells you the same thing (emphasis added):

Please note that the existence of the above limit does not mean that the function $f$ has a second derivative. The limit above just gives a possibility for calculating the second derivative but does not provide a definition.