急.求极限 lim x趋向于π/2,(sinx)的(1/cosx)^2求极限 lim x趋向于π/2,(sinx)的(1/cosx^2)次方

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急.求极限limx趋向于π/2,(sinx)的(1/cosx)^2求极限limx趋向于π/2,(sinx)的(1/cosx^2)次方急.求极限limx趋向于π/2,(sinx)的(1/cosx)^2求

急.求极限 lim x趋向于π/2,(sinx)的(1/cosx)^2求极限 lim x趋向于π/2,(sinx)的(1/cosx^2)次方
急.求极限 lim x趋向于π/2,(sinx)的(1/cosx)^2
求极限 lim x趋向于π/2,(sinx)的(1/cosx^2)次方

急.求极限 lim x趋向于π/2,(sinx)的(1/cosx)^2求极限 lim x趋向于π/2,(sinx)的(1/cosx^2)次方
设 f(x) = (sinx) ^ (1/cos²x ),ln f(x) = (1/cos²x ) ln(sinx)
lim (x->π/2) ln f(x) = lim (x->π/2) ln(sinx) / cos²x
= lim(t->0) ln(cost) / sin²t 令 t = π/2 - x,t->0
= lim(t->0) ln(1+ cost-1) / t² 等价无穷小代换 ln(1+ cost-1) cost - 1 -t² /2
= lim(t->0) (-1/2) t² / t² = -1/2
原式 = e ^ (-1/2)

lim x趋向于π/2,(sinx)的(1/cosx^2)次方
= lim x趋向于π/2,(1+sinx-1)的(1/cosx^2)次方
=e^(lim x趋向于π/2,(sinx-1)/(cosx)^2)
=e^{lim x趋向于π/2,(cosx)/[2(cosx)*(-sinx)]}
=e^(-1/2)

先取对数,之后用到诺比达法则,我算得答案是e的(-1/2)次方。