Question

一道高数题求助在线等

Answer 1

f(x)
=x^α.cos(1/x^β) ; x>0
=0 ; x≤0
lim(x->0) f(x)
=lim(x->0) x^α.cos(1/x^β)
=0
=>α>0
x≠0
f'(x)
=x^(α-1).cos(1/x^β) +x^α. [-sin(1/x^β) ]. [ -β/x^(β+1) ]
=x^(α-1).cos(1/x^β) +βx^(α-β-1). [sin(1/x^β) ]
f'(0)
lim(h->0) [h^α.cos(1/h^β) - f(0) ]/h
=lim(h->0) h^(α-1).cos(1/h^β)
=0
=>
α-1 >0
α>1
lim(x->0) f'(x)
=lim(x->0) { x^(α-1).cos(1/x^β) +βx^(α-β-1). [sin(1/x^β) ] }
=0
=>
α-1 >0 and α-β-1 >0
α >1 and α-β >1
=>
α-β >1
ans: A

Answer 2

简单计算一下即可，答案如图所示