一道高数题，如图。这题答案为什么是这样做的?

可以用Taylor展开式，展开到几阶那是试出来的；也可以试试L'Hospital法则，再用Taylor展开式，会更简单。

x->0
sinx = x- (1/6)x^3 +(1/120)x^5+o(x^5)
asinx =ax -(1/6)ax^3 +(1/120)ax^5 +o(x^5)
sin2x =2x- (4/3)x^3 + (4/15)x^5 +o(x^5)
(b/2)sin2x=bx -(2/3)bx^3 +(2/15)bx^5 +o(x^5)
x- asinx -(b/2)sin2x
=x -[ax -(1/6)ax^3 +(1/120)ax^5 +o(x^5)] -[bx -(2/3)bx^3 +(2/15)bx^5 +o(x^5)]
= (1-a-b)x + [(1/6)a + (2/3)b]x^3 + [ -(1/120)a - (2/15)b] x^5 +o(x^5)
最高阶
1-a-b=0 (1) and
(1/6)a + (2/3)b =0 (2)
sub (1) into (2)
(1/6)a + (2/3)b =0

(1/6)a +( 2/3)(1-a) = 0
(1/2)a= 2/3
a=4/3
from (1)
1-a-b=0
1-4/3-b=0
b= -1/3
(a,b)= (4/3, -1/3)
x->0
x-(a+bcosx).sinx
=x -[ 4/3 -(1/3)cosx )sinx
=[ -(1/120)(4/3) - (2/15)(-1/3) ] x^5 +o(x^5)
=( -1/90 + 2/45) x^5 +o(x^5)
=(1/30)x^5 +o(x^5)