x→0,lim(1-cosx)[x-ln(1+tanx)]/sinx^4的极限

首先用等价无穷小代换,(1-cosx)换成1/2x^2,sinx^4换成x^4
lim(1-cosx)[x-ln(1+tanx)]/sinx^4
=lim(1/2)x^2[x-ln(1+tanx)]/x^4
=lim(1/2)[x-ln(1+tanx)]/x^2
洛必达法则
=lim(1/2)*[1-(secx)^2/(1+tanx) ]/(2x)
=lim(1/2)*[1+tanx-(secx)^2]/[2x(1+tanx)]
=lim(1/2)*[tanx-(tanx)^2]/[2x(1+tanx)]
再用等价无穷小代换,tanx可换为x
=lim(1/2)*(1-tanx)/[2(1+tanx)]
=1/4
其中：[x-ln(1+tanx)]'=1-(secx)^2/(1+tanx)