lim n→∞(3n2+cn+1/an2+bn-4n)=5,求常数a、b、c的值是 3n2+cn+1/an2+bn -4n 4n单独做一个减数，而不是在分母上做减数

lim n→∞(3n2+cn+1/an2+bn-4n)=5,求常数a、b、c的值是 3n2+cn+1/an2+bn -4n 4n单独做一个减数，而不是在分母上做减数
lim n→∞(3n2+cn+1/an2+bn-4n)=5,求常数a、b、c的值
是 3n2+cn+1/an2+bn -4n 4n单独做一个减数，而不是在分母上做减数

lim n→∞(3n2+cn+1/an2+bn-4n)=5,求常数a、b、c的值是 3n2+cn+1/an2+bn -4n 4n单独做一个减数，而不是在分母上做减数
上下除以n²
(3+c/n+1/n²)/(a+b/n-4/n²)
n→∞
所以分母有n的极限都是0
所以极限=3/a=5
a=3/5
无法求出常数b和c

1、
f(x)=4[1-cos2(π/4+x)]/2-2√3cos2x-1
=2[1-cos(π/2+2x)]-2√3cos2x-1
=2+2sin2x-2√3cos2x-1
=4(sin2x*1/2-cos2x*√3/2)+1
=4(sin2xcosπ/3-cos2xsinπ/3)+1
=4sin(2x-π/3)-1
π/4<=x<=π/2...

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1、
f(x)=4[1-cos2(π/4+x)]/2-2√3cos2x-1
=2[1-cos(π/2+2x)]-2√3cos2x-1
=2+2sin2x-2√3cos2x-1
=4(sin2x*1/2-cos2x*√3/2)+1
=4(sin2xcosπ/3-cos2xsinπ/3)+1
=4sin(2x-π/3)-1
π/4<=x<=π/2
π/6<=2x-π/3<=2π/3
所以1/2<=sin(2x-π/3)<=1
1<=4sin(2x-π/3)-1<=3
所以值域[1,3]
2、
sinx增区间是(2kπ-π/2,2kπ+π/2)
π/6<=2x-π/3<=2π/3
所以这里是(π/6,π/2)
π/6<2x-π/3<π/2
π/2<2x<5π/6
π/4同理得到减区间
所以
增区间(π/4,5π/12)
减区间(5π/12,π/2)

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