lim
= (1/2) lim
= (1/2) { lim
= (1/2) [f'(x0) + f'(x0)] = f'(x0), f(x) 在 x = x0 处可导。
已知h趋于0时f(x0+h)-f(x0-h)/2h存在，问f(x0)是否可导
当 f(x) 连续时，
lim[f(x0+h)-f(x0-h)]/(2h)
= (1/2) lim[f(x0+h)-f(x0)+f(x0)-f(x0-h)]/h
= (1/2) { lim[f(x0+h)-f(x0)]/h + lim [f(x0-h)-f(x0)]/(-h) }
= (1/2) [f'(x0) + f'(x0)] = f'(x0), f(x) 在 x = x0 处可导。
lim
= (1/2) lim
= (1/2) { lim
= (1/2) [f'(x0) + f'(x0)] = f'(x0), f(x) 在 x = x0 处可导。
