Question

高数，微积分，求大神

Answer 1

详细解答

追答

望采纳

Answer 2

f(x)=1-cosxcos2xcos3x

积化和差:

f(x)=1-cosx[cos5x+cosx]/2

=1-(cosxcos5x+cos^2x)/2

=1-(cos4x+cos6x+cos2x+1)/4

=3/4-(cos2x+cos4x+cos6x)/4

f'(x)=(2sin2x+4sin4x+6sin6x)/4

f''(x)=(4cos2x+16cos4x+36cos6x)/4

x->0时,

lim(f(x)/(ax^n)=lim(4cos2x+16cos4x+36cos6x)/4 /(ax^n)''

=lim(4+16+36)/[4 (a(n)(n-1)x^(n-2)]

=lim14/(an(n-1)x^(n-2))

n-2=0   n=2

14/[a*2*(2-1)]=1

14=2a

a=7

自已再算一下,顺便说一下,楼上的算法;在加法时用等价无穷小替换,是错误的,答案碰巧对,不会给分的.不过他的方法有意思:

-f(x)=-1+cosxcos2xcos3x

-f(x)~ln(1+(-1+cosxcos2xcos3))

-f(x)~ln(cosxcos2xcos3x)=lncosx+lncos2x+lncos3x

替换后得:lim-[(ln(cosx)+ln(cos2x)+ln(cos3)]/(ax^n)

=-lim(-sinx/cosx-2sin2x/cos2x-3sin3x/cos3x)/(anx^(n-1)

=-lim(-sec^2x-4sec^2(2x)-9sec^2(3x))/(an(n-1)x^(n-2))

=-lim(-1-4-9)/(an(n-1)x^(n-2)

n-2=0   n=2

a*2*1=14   a=7