It is proved that f (0) = LIM (x - > 0) [f (x) + F (- x)] / 2

It is proved that f (0) = LIM (x - > 0) [f (x) + F (- x)] / 2

The function is continuous at x = 0, otherwise it can't be proved. If you have this condition, you can prove it. Let the formula on the right subtract two half f (0). Take the limit respectively. From the continuity, the result is 0

If f is differentiable at x = a, find LIM (x approaches 0) (f (a + H) - f (A-H)) / 2H

If f (x) is differentiable, find the value of LIM [f (x) - f (X-H)] / h when h approaches zero
This is a one-sided reciprocal problem, divided into left and right derivatives

Let H = - t, then when h → 0 -, t → 0+
So the original formula = Lim [t → 0 +] [f (x) - f (x + T)] / (- t)
=lim【t→0+】[f(x+t)-f(x)]/t
=F '+ (x), that is, the right derivative of F (x) at point X!