第4个回答 2021-12-18
x->0
sin4x = 4x -(1/6)(4x)^3 +o(x^3) = 2x -(8/3)x^3 +o(x^3)
(1/2)sin4x = 2x -(4/3)x^3 +o(x^3)
2x-(1/2)(sin4x) =(4/3)x^3 +o(x^3)
//
lim(x->0) [ 1/(sinx)^2 - (cosx)^2/x^2]
=lim(x->0) [x^2-(sinx.cosx)^2] /[x^2.(sinx)^2]
=lim(x->0) [x^2-(sinx.cosx)^2] /x^4
=lim(x->0) [x^2-(1/4)(sin2x)^2] /x^4
洛必达
=lim(x->0) [2x-(1/2)(sin4x)] /(4x^3)
=lim(x->0) (4/3)x^3 /(4x^3)
=1/3