Question

导数的证明证明f（x）在x0处的导数=lim（n趋向于无穷大）（（f（x0+an）-f（x0-bn））/（an+bn））

Answer 1

对任意 ε>0,由条件,因

lim(n→inf.)[f(x0+an) - f(x0)]/an = f'(x0),

lim(n→inf.)[f(x0+bn) - f(x0)]/bn = f'(x0),

存在正整数 N,使当 n>N 时,有

|[f(x0+an) - f(x0)]/an - f'(x0)| < ε,|[f(x0+bn) - f(x0)]/bn - f'(x0)| < ε,

此时

|[f(x0+an) - f(x0-bn)]/(an+bn) - f'(x0)|

= |{[f(x0+an) - f(x0)]/an - f'(x0)}[an/(an+bn)] - {[f(x0+bn) - f(x0)]/bn - f'(x0)}[bn/(an+bn)]|