用洛必达法则求极限
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而，lim(x→π/4)(tan2x)ln(tanx)=lim(x→π/4)(sin2x)*lim(x→π/4)ln(tanx)/cos2x=lim(x→π/4)ln(tanx)/cos2x，属“0/0”型，用洛必达法则，
∴lim(x→π/4)(tan2x)ln(tanx)=lim(x→π/4)(sin2x)*lim(x→π/4)ln(tanx)/cos2x=-lim(x→π/4)1/sin²2x=-1，
∴原式=e^(-1)。
供参考。
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