Question

(x+sinx)*(x-sinx)求导

这步洛必达怎么弄出来的啊？我感觉前面系数是1/2

Answer 1

sinx= x-(1/6)x^3+o(x^3)
x-sinx=(1/6)x^3+o(x^3)
lim(x->0) (x+sinx)(x-sinx)/x^4
=lim(x->0) 2x.(x-sinx)/x^4
=lim(x->0) 2x.(1/6)x^3/x^4
=1/3

or
y=(x+sinx)(x-sinx)
y'= 2x -sin2x
x->0
sin2x = 2x-(1/6)(2x)^3 +o(x^3) = 2x - (4/3)x^3+o(x^3)
2x-sin2x= (4/3)x^3+o(x^3)
lim(x->0) (x+sinx)(x-sinx)/x^4

洛必达
=lim(x->0) (2x-sin2x)/(4x^3)
=lim(x->0) (4/3)x^3/(4x^3)
=1/3
根本上是一样！

追问

我想问问洛必达哪错了

追答

lim(x->0) (x+sinx)(x-sinx)/x^4

洛必达

=lim(x->0) [ (x+sinx)(1-cosx) +(x-sinx)(1+cosx) ]/(4x^3)

=2lim(x->0) (x-sinx)/(4x^3) : 这个不对，我不知道怎样来！

追问

我悟了大佬，那个1-cosx=1/2*x^2, (x+sinx)(1-cosx)就=x^3=6（x-sinx），人麻了都

Answer 2

=lim(x->0) 2x.(x-sinx)/x^4 =lim(x->0) 2x.(1/6)x^3/x^4 =1/3 or y=(x+sinx)(x-sinx)y&9;= 2x -sin2x x->0 sin2x = 2x-(1/6)(2x)^3 +o(x^3) = 2x - (4/3)x^3+o(x^3)2x...

Answer 3

=lim(x->0) 2x.(x-sinx)/x^4 =lim(x->0) 2x.(1/6)x^3/x^4 =1/3 or y=(x+sinx)(x-sinx)y&9;= 2x -sin2x x->0 sin2x = 2x-(1/6)(2x)^3 +o(x^3) = 2x - (4/3)x^3+o(x^3)2x...