Question

求解第4题，要有详细解法

Answer 1

解：
1+acos2x+bcos4x
=-a+acos2x+b+bcos4x+a-b+1
=-a(1-cos2x)+b(1+cos4x)+(a-b+1)
=-a·2sin²x+b·2cos²2x +a-b+1
=-2asin²x+2b(1-2sin²x)²+a-b+1
=-2asin²x+2b(1-4sin²x+4sin⁴x)+a-b+1
=8bsin⁴x-(2a+8b)sin²x+a+b+1
lim (1+acos2x+bcos4x)/x⁴
x→0
=lim [8bsin⁴x-(2a+8b)sin²x+a+b+1]/x⁴
x→0
=lim [8bx⁴-(2a+8b)x²+a+b+1]/x⁴
x→0
=lim [8b -(2a+8b)/x²+ (a+b+1)/x⁴]
x→0
要极限存在
2a+8b=0
a+b+1=0
解得a=-4/3，b=⅓
lim (1+acos2x+bcos4x)/x⁴=8b=8·⅓=8/3
x→0
常数a的值为-4/3，b的值为⅓，极限值为8/3

Answer 2

lim(x->0) (1+acos2x+bcos4x)/x^4

x->0
cosx ~ 1-(1/2)x^2+(1/24)x^4
acos2x ~ a[ 1-2x^2+(2/3)x^4 ]
bcos4x ~ b[ 1-8x^2+(32/3)x^4 ]

1+acos2x+bcos4x ~(1+a+b) +(-2a-8b)x^2 +( (2/3)a + (32/3)b)x^4
lim(x->0) (1+acos2x+bcos4x)/x^4 存在

=>
1+a+b =0 (1) and
-2a-8b=0 (2)

2(1) +(2)
2-6b=0
b=1/3

from (1)
1+a+1/3=0
a=-4/3

(2/3)a + (32/3)b
= (2/3)(-4/3) + (32/3)(1/3)
=-8/9+ 32/9
=24/9
=8/3

lim(x->0) (1+acos2x+bcos4x)/x^4
=lim(x->0) (8/3)x^4/x^4
=8/3

Answer 3