Question

一道高数题设f(x)=x^a*cos1/x(x≠0）,f(x)=0(x=0),其导函数在x=0处连续,求a的取值范围

Answer 1

导函数连续则函数必须连续

即lim[x->0]f(x) = f(0) = 0

而cos(1/x)是有界函数,a>0时x^a->0,则lim[x->0]x^a * cos(1/x) = 0,即a>0时f(x)都连续.

而f'(x) = a x^(a-1) * cos(1/x) + x^a * [-sin(1/x)] * (-1/x²) = a x^(a-1) * cos(1/x) + x^(a-2) * sin(1/x)

导函数连续则lim[x->0]f'(x) = f'(0)

f'(0) = lim[Δx->0][f(Δx)-f(0)]/Δx = lim[Δx->0][(Δx)^a * cos(1/Δx) - 0]/Δx = 0 (a>1)

lim[x->0]f'(x) = 0

即lim[x->0] [a x^(a-1) * cos(1/x) + x^(a-2) * sin(1/x)] = 0

这要求a-1>0且a-2>0

则a>2