Question

这道高数题怎么做？

Answer 1

f(x) = x- (a+bcosx)sinx = x-asinx -(b/2)sin2x

=>f(0) = 0

f'(x) = 1 - acosx - bcos2x

f'(0)=0

1-a-b =0

a+b=1                                               (1)

f''(x) =  asinx + 2bsin2x

f''(0) =0

f'''(x) =  acosx + 4bcos2x

f'''(0)=0

a+4b=0                                              (2)

(1)-(2)

-3b=1

b=-1/3

from (1)

a+b=1 

a-1/3=1

a= 4/3

ie

(a,b) = (4/3 -1/3)

f'''(x) =  acosx + 4bcos2x =(4/3)cosx -(4/3)cos2x

f''''(x) = -(4/3)sinx +(8/3)sin2x

f^(5)(x) =-(4/3)cosx +(16/3)cos2x

f^(5)(0)  = -4/3 +16/3 = 4

f^(5)(0)/5! = 1/30

lim(x->0) f(x)/x^5

=lim(x->0) (1/30)x^5/x^5

=1/30

            

Answer 2

Answer 3