tan t-sin x 解.liml (1+tanx lim 1+ x→0(1+sinx 1+sin x 性x-512X 1+sinx a+sinx)x tanx-5Inx tan x-sinx tanr-sinr =11m x→0 1+sin x sin x(1-cosx 2x25i2 (2)求下列极限 解.当x→1时,ln(1+x-1)~x-1, arcsin2yx2-1~2yx2-1.按照等价无穷小 代换lim n(1+x-1) -lim =lim x- arcsin2x2-122x2-122x+12V2 x+(x' 5m.r=lim/sin'x-xcos'x 1 coS-x lin cotx=lim x SIn x =lim 1-(x2+1)cos2x 2x cos x+2(x+1)cos xsin x =lim x→0 2x- cos sinx lin -2x cos- x+ sin 2x +lim 4 lim. x+4.x cos x sin x 2 cos 2. x, 1 12x 2 2 cos-x+2 cos 21 1 1 4 cosxsin x-4sin 2x 1 1 lir lir 12x 24x 2 sin 2x =llm x→024x (3)求下列极限⑵求下列极限 ① ② ⑶求下列极限 ①