Question

第二十题怎么做呢？

Answer 1

f(x)
= x^2 + x^k . sin(1/x) ; x≠0
=0 ; x=0
lim(x->0) f(x) = 0
lim(x->0) [x^2 + x^k . sin(1/凳答x)] =0
=> k >0
k>0, x=0 , f(x) 连续

f'(0)
=lim(h->0) [f(h)- f(0) ]/h
=lim(h->0) [h^2 + h^k . sin(1/h) ]/h
=lim(h->0) [h + h^(k-1) . sin(1/h) ]
=0
k-1>喊粗顷0
k>1
ie
k> 1 , x= 0, f(x) 可导
f''(0)
=lim(h->0) [ f'(h)-f'(0) ]/h
=lim(h->1) [2h + k.h^(k-1) . sin(1/h) - h^(k-2) . cos(1/h) ]/h
=lim(h->郑陆1) [2 + k.h^(k-2) . sin(1/h) - h^(k-3) . cos(1/h) ]
=2 (存在)
k-2 >0 and k-3>0
k>3
ie
k>3 , f''(0) = 2

追问

为什么k-2大于0就存在导数呢

追答

f(x)
= x^2 + x^k . sin(1/x) ; x≠0
=0 ; x=0
lim(x->0) f(x) = 0
lim(x->0) [x^2 + x^k . sin(1/x)] =0
=> k >0
k>0, x=0 , f(x) 连续
f'(0)
=lim(h->0) [f(h)- f(0) ]/h
=lim(h->0) [h^2 + h^k . sin(1/h) ]/h
=lim(h->0) [h + h^(k-1) . sin(1/h) ]
=0
k-1>0
k>1
ie
k> 1 , x= 0, f(x) 可导
f''(0)
=lim(h->0) [ f'(h)-f'(0) ]/h
=lim(h->0) [2h + k.h^(k-1) . sin(1/h) - h^(k-2) . cos(1/h) ]/h
=lim(h->0) [2 + k.h^(k-2) . sin(1/h) - h^(k-3) . cos(1/h) ]
=2 (存在)
k-2 >0 and k-3>0
k>3
ie
k>3 , f''(0) = 2