y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x.y=lim (n → ∞)

就是爱草2022-10-04 11:39:541条回答

y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x.y=lim (n → ∞) cos^2n (arctanx).
y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x.
y=lim (n → ∞) cos^2n (arctanx).
y=lim (cos2x)^(1+cot^2x) (这道题用ln公式做,我想知道用的哪个ln公式)
我怕我分析不出来.

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bestyaose 共回答了22个问题 | 采纳率86.4%
1.y=lim (x → 0) ( √1+xsinx - √cosx) / arcsin^2x
=lim (x → 0) {[(sinx+cosx)/2√(1+xsinx)+sinx/2√cosx]}/[2arcsinx/√(1-x²)
=lim(x → 0) [1/2+0]/(2arcsinx)
=lim(x → 0) 1/[4arc sinx]
=∞
2.y=lim (n → ∞) cos^2n (arctanx).
=lim (n → ∞) (1/2)[1+cos2n(arctanx)]
=lim (n → ∞) (1/2)*[1+cosnπ]
=(1/2)*(1±1)
=0或1
3.y=lim(x → π/2) (cos2x)^(1+cot^2x)
=lim(x → π/2) e^[(1+cot²x)lncos²x)]
=lim(x → π/2) e^[2(lncosx)/cos²x)]
=lim(x → π/2) e^[(-2sinx/cosx)/(-2cosxsinx)]
=lim(x → π/2) e^(1/cos²x)
=lim(x → π/2) e^[2/(cos2x+1)]
=e^[2/(-1+1)]
=e^∞
=∞
1年前

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