Question

lim(x→0)(cos x)∧(1/sin 2x)＝?

Answer 1

f(x) = (cos x)^(1/sin 2x), lnf(x) = 1/(sin2x) ln cosx = ln(cosx) / sin2x
lim(x->0) lnf(x)
= lim(x->0) ln(cosx) / sin2x ln(1 + cosx-1) ~ cosx -1
= lim(x->0) (cosx -1) / (2x) cosx -1 ~ -x²/2
= lim(x->0) (-x²/2) / (2x )
= 0
原式 = e^0 = 1

Answer 2

令y=cosx^(1/sin2x),则有：
lny=lncosx/sin2x
lim(x→0）lny
=lim（x→0）lncosx/sin2x 应用罗必塔法则可得到：
=lim(x→0）（-sinx/cosx)/(2cos2x)=0
所以：
lim(x-0)y=1.