来自田绍槐的问题
求Lim(x→π/2)(sinx)^tanx,不用洛必达法则如何解?
求Lim(x→π/2)(sinx)^tanx,不用洛必达法则如何解?
1回答
2020-12-21 00:12
求Lim(x→π/2)(sinx)^tanx,不用洛必达法则如何解?
求Lim(x→π/2)(sinx)^tanx,不用洛必达法则如何解?
=e^lim(x→π/2)ln[(sinx)^tanx]
=e^lim(x→π/2)tanx·ln(sinx)
=e^lim(x→π/2)ln[1+(-1+sinx)]/tan(π/2-x)
=e^lim(x→π/2)(-1+sinx)/(π/2-x)【注意π/2-x→0,则tan(π/2-x)~(π/2-x);t→0时ln[1+t]~t】
=e^lim(x→π/2)(1-sin²x)/[(π/2-x)(-1-sinx)]
=e^lim(x→π/2)-(cos²x)/[(π/2-x)(1+sinx)]
=e^(-1/2)·lim(x→π/2)(cos²x)/(π/2-x)
=e^(-1/2)·lim(x→π/2)sin²(π/2-x)/(π/2-x)
=e^(-1/2)·lim(x→π/2)(π/2-x)²/(π/2-x)
=e^(-1/2)·lim(x→π/2)(π/2-x)
=e^0
=1
【不用洛必达法还真挺绕呐】