当x趋向于π/2 时,求tanx^sin2x的极限
1个回答

y=tanx^sin2x

lny=sin2xlntanx

lim[x-->π/2]sin2xlntanx

=lim[x-->π/2]lntanx/csc2x (∞/∞型,用洛必达法则）

=lim[x-->π/2](secx)^2/(-2csc2xcot2xtanx)

=lim[x-->π/2](sin2x)^2cosx/[-2cos2xsinx(cosx)^2]

=lim[x-->π/2](sin2x)^2/(2cosx)

=lim[x-->π/2]4sin2xcos2x/(-2sinx)

=0

∴lim[x-->π/2]tanx^sin2x=e^0=1

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