Question

求极限limx→0 (cosx)^1/sin^2x

求极限limx→0 (cosx)^(1/sin^2x)

Answer 1

(cosx)^(1/sin^2x)=eln( (cosx)^(1/sin^2x))=e((1/sin^2x)lncosx)

里面的是0/0型，上下求一次导数，就出来了
我解出来的是：e的(1/2)次方

Answer 2

取对数，再用洛必达法则。
ln原式=lim(x→0)ln(cosx)/sin^2(x)=lim(x→0)(-sinx/cosx)/(2sinxcosx)=lim(x→0)-1/(2cos^2(x))=-1/2
所以原式=e^(-1/2)

Answer 3

设sinx^2=t
原式=[(1-t)^1/2]^(1/t)=(1-t)^1/2t=e^(-1/2)

Answer 4

=e^ lim (1/sin²x)·ln cosx
=e^ lim (cosx -1)/x²
=e^ lim -(1/2)x²/x²
=e^ -(1/2)

Answer 5

lim(x->0 )(cosx)^(1/sinx^2)= lim(x->0 )(1-1/2*x^2)^(1/x^2)
= lim(x->0 )(1-1/2*x^2)^(-1/2*1/(-1/2*x^2))
= lim(x->0 )(1-1/2*x^2)^( 1/(-1/2*x^2)*( -1/2))
=e^( -1/2)