Question

求极限(sinx)^tanx,当x趋向与π/2的时候

=lim(x->Л/2)[(1/sinx)*cosx]/[-(sinx)^2-(cosx)^2/(sinx)^2]
=lim(x->Л/2)1/2*sin2x
=0

这几步骤是怎么来的？谢谢

Answer 1

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