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】