lim x->0 x^a sin(1/x)=f(0)
sin(無窮）屬于【0,1】
所以 limx->0 x^a=0
a>0
f(x)連續(xù)
2）
limx->0(x^a sin(1/x))’=f'(0)
x^a sin(1/x)'=ax^(a-1)sin(1/x)-cos(1/x)*(x)^(-2)x^a
=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
f'(0)=lim h->0 [f(h)-f(0)]/h
=limh->0 h^a sin(1/h)/h
=limh->0 h^(a-1) sin(1/h)
所以我們需要滿足
lim x->0 ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)=limh->0 h^(a-1)sin(1/h)
如果a