这题怎么解

x->0
分子
cosx -cos2x = [1- (1/2)x^2] - [ 1- (1/2)(2x)^2] +o(x^2)= (3/2)x^2 +o(x^2)
tanx = x +o(x^2)
ln(1+tanx) = ln(1+x+o(x^2))= x-(1/2)x^2+o(x^2)
x-ln(1+tanx) = (1/2)x^2+o(x^2)
(cosx -cos2x) [ x- ln(1+tanx)] = (3/4)x^4 +o(x^4)
分母
sinx = x - (1/6)x^3 +o(x^4)
cos(sinx)
=cos[x - (1/6)x^3 +o(x^4)]
=1- (1/2)[x - (1/6)x^3]^2 + (1/24)[x - (1/6)x^3]^4 +o(x^4)
=1- (1/2)[x^2 - (1/3)x^4 +o(x^4)] + (1/24)[x^4+o(x^4)] +o(x^4)
=1- (1/2)x^2 + ( 1/6 +1/24)x^4 +o(x^4)
=1- (1/2)x^2 + (5/24)x^4 +o(x^4)
cosx =1- (1/2)x^2 + (1/24)x^4 +o(x^4)
cos(sinx) -cosx = (5/24 -1/24)x^4 +o(x^4) = (1/6)x^4+o(x^4)
//
lim(x->0+) (cosx -cos2x) [ x- ln(1+tanx)]/[cos(sinx) -cosx]
=lim(x->0+) (3/4)x^4/[(1/6)x^4]
=9/2