Question

求Lim(x→π/2)(sinx)^tanx，不用洛必达法则如何解？

Answer 1

=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】

Answer 2

令(sinx)^tanx=y
lny=tanx ln sinx=ln(sinx)/(1/tanx)
然后令x趋向于0,y=1