Are these two different derivative definitions equal?

Question

I have a question which asks me if the following two definitions of a derivative is equal. So I know the following equation,

$f'(x_0) = \lim_{h \to 0} \dfrac{f(x_0+h) - f(x_0)}{h} $

as we went through it when I learnt this and how to get the derivative using limits as h goes to $0$ but I don't get the following equation.

$g'(x_0) = \lim_{h \to 0} \dfrac{g(x_0+h) - g(x_0-h)}{2h} $

The question is asking me knowing that we defined a derivative of $\ f$ at $\ x_0$, is the following definition suggested by someone as equivalent and if so why? How would I go about answering this? Super confused about where to start?

Answer 1

Consider the function $g(x) = |x|$ and $x_0 =0$ or the function $g(x) = 1/x^2$ and $x_0 = 0$ (not in the domain). The second "definition" would imply $g'(0) = 0$.

If the second "definition" were true, then the value of the function at $x_0$ would be irrelevant - hence the "derivative" at $x_0$ might exist even if the function were not continuous, or not even defined at $x_0$.

Answer 2

Remember the derivative is the slope of the tangent line. This is obtained by taking a limit of slopes of secant lines. So, if you consider the points $(x-h, f(x-h))$ and $(x+h, f(x+h))$, using the slope formula gives $$ m = \frac{f(x+h) - f(x-h)}{(x+h) - (x-h)} $$ which of course cleans up as $$ m = \frac{f(x+h) - f(x-h)}{2h}. $$ Now pass $h \to 0.$

In short, yes, the definitions are equivalent.