Question

两道大学数学求极限问题

请给出详细过程，好的加分哦！

Answer 1

1）原式=lim(x-sinx)/(1-cosx) lim1/ln(1+x) 拆分成两部分的积
lim(1-cosx)/sinx lim1/ln(1+x) 前一部分用洛必达法则
=limsinx/cosx lim1/ln(1+x) 前一部分再用洛必达法则
=limtanx/ln(1+x) 化简
=lim(1+x)/cos^2(x) 再用洛必达法则
=(1+0)/1
=1 (x->0)

2）原式=lim exp { ln(1/cosx)^[1/x(x-1)] }
=lim exp { [1/x(x-1)] ln(1/cosx) }
=lim exp { ln(1/cosx) / (x^2-x) }
=lim exp { [cosx /(-cos^2(x)) * (-sinx)] / (2x-1)} 用洛必达法则
=lim exp [tanx/(2x-1)]
=lim exp 0/(-1) (x->0)
=1

Answer 2

解：
(1) 此题用等价无穷小来解：
(x→0)lim(x-sinx)/[(1-cosx)*ln(1+x)]
= (x→0)lim[x-(x-x³/3!+0(x³))]/{[1-(1-x²/2!+0(x²))]*[x+0(x)]
= (x→0)lim[x³/3!+0(x³)]/[(x²/2!+0(x²))*(x+0(x))]
= (x→0)lim[x³/6+0(x³)]/[(x³/2+0(x³)] = 1/3
(2) (x→0)lim(1/cosx)^[1/(x(x-1))]
= (x→0)lim(secx)^[1/(x(x-1))]
= (x→0)lim(1+tan²x)^[1/2*1/(x(x-1))]
= (x→0)lim(1+tan²x)^[(1/tan²x)* tan²x/(2x(x-1))]
= (x→0)lim[(1+tan²x)^(1/tan²x)]^[tan²x/(2x(x-1))]
= e^0 = 1

追问

我感觉楼上的第一题才是对的，不知在下看法如何？谢谢指导