求lim(x→0) (√1-cosx^2)/(1-cosx),还有题lim(x→0) (x-xcosx)/(tanx-sinx),不用洛必达法则,怎么求这两道题的极限呢?

来源：学生作业帮助网编辑：作业帮时间：2024/04/26 12:27:40

求lim(x→0) (√1-cosx^2)/(1-cosx),还有题lim(x→0) (x-xcosx)/(tanx-sinx),不用洛必达法则,怎么求这两道题的极限呢?
求lim(x→0) (√1-cosx^2)/(1-cosx),还有题lim(x→0) (x-xcosx)/(tanx-sinx),
不用洛必达法则,怎么求这两道题的极限呢?

求lim(x→0) (√1-cosx^2)/(1-cosx),还有题lim(x→0) (x-xcosx)/(tanx-sinx),不用洛必达法则,怎么求这两道题的极限呢?
（1）lim(x->0)[√(1-cos(x²))/(1-cosx)]=lim(x->0)[√(2sin²(x²/2))/(2sin²(x/2))] (应用半角公式)
=√2lim(x->0)[(sin(x²/2)/(x²/2))((x/2)/sin(x/2))²]
=√2{lim(x->0)[sin(x²/2)/(x²/2)]}*{lim(x->0)[(x/2)/sin(x/2)]}²
=√2*1*1² (应用重要极限lim(z->0)(sinz/z)=1)
=√2
（2）lim(x->0)[(x-xcosx)/(tanx-sinx)]=lim(x->0)[x(1-cosx)/(sinx/cosx-sinx)]
=lim(x->0)[x(1-cosx)cosx/sinx(1-cosx)]
=lim(x->0)[(x/sinx)cosx]
=[lim(x->0)(x/sinx)]*[lim(x->0)(cosx)]
=1*1 (应用重要极限lim(z->0)(sinz/z)=1)
=1

可以用泰勒级数展开cosx=1-x^2+o(x^2)【这表示x^2的无穷小】然后代入上式就出来了啊

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