我只想说,你给出的解题过程是错的,而且是没有道理的
x→π/2
lim (sinx)^tanx
换元：t=π/2-x,x=π/2-t
=lim(t→0) [sin(π/2-t)]^tan(π/2-t)
=lim (cost)^cott
=lim e^ln (cost)^cott
根据复合函数的极限运算：lim(x→x0) f(g(x))=f(lim(x→x0) g(x))
=e^ lim ln (cost)^cott
现在考虑
lim ln (cost)^cott
=lim cott * ln cost
=lim ln cost / tant
=lim ln(1+cost-1) / tant
利用等价无穷小：ln(1+x)~x,tanx~x
=lim cost-1 / x
再利用等价无穷小：1-cosx~x^2/2
=-lim x^2 / 2x
=-lim x/2
=0
因此
lim(x→π/2) (sinx)^tanx
=e^ lim(t→0) ln (cost)^cott
=e^0
=1
有不懂欢迎追问