Question

使用洛必达法则求导 lim(x→pi/2-0):(tanx)^(2x-pi) pi=3.1415926....... 求过程 谢谢.

Answer 1

解：lim(x→π/2-0) (tanx)^(2x-π)
=e^ lim(x→π/2-0) (2x-π)ln(tanx)
=e^ lim(x→π/2-0) ln(tanx)/[1/(2x-π)] ∞/∞型，用罗比达法则
=e^ lim(x→π/2-0) 1/(tanx)*sec^2x/[-2/(2x-π)^2]
=e^ lim(x→π/2-0) -(2x-π)^2/sin(2x)
=e^ lim(x→π/2-0) -(2x-π)^2/sin(2x-π+π)
=e^ lim(x→π/2-0) (2x-π)^2/sin(2x-π)
=e^ lim(x→π/2-0) (2x-π)^2/(2x-π)
=e^(2*π/2-π)
=e^0=1

追问

=e^ lim(x→π/2-0) 1/(tanx)*sec^2x/[-2/(2x-π)^2]
=e^ lim(x→π/2-0) -(2x-π)^2/sin(2x)
这步是怎么转化的

追答

1/(tanx)*sec^2x/[-2/(2x-π)^2]
=cosx/sinx*1/(cosx)^2*[-(2x-π)^2]/2
=1/(sinxcosx)*[-(2x-π)^2]/2
=-(2x-π)^2/(2sinxcosx)
=-(2x-π)^2/sin(2x)