精品偷拍一区二区三区,亚洲精品永久 码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲

  • <center id="usuqs"></center>
  • 
    
  • 求lim(x→0) (√1-cosx^2)/(1-cosx),還有題lim(x→0) (x-xcosx)/(tanx-sinx),

    求lim(x→0) (√1-cosx^2)/(1-cosx),還有題lim(x→0) (x-xcosx)/(tanx-sinx),
    不用洛必達(dá)法則,怎么求這兩道題的極限呢?
    數(shù)學(xué)人氣:473 ℃時(shí)間:2019-09-24 04:38:56
    優(yōu)質(zhì)解答
    (1)lim(x->0)[√(1-cos(x²))/(1-cosx)]=lim(x->0)[√(2sin²(x²/2))/(2sin²(x/2))] (應(yīng)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 (應(yīng)用重要極限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 (應(yīng)用重要極限lim(z->0)(sinz/z)=1)
    =1
    我來回答
    類似推薦
    請使用1024x768 IE6.0或更高版本瀏覽器瀏覽本站點(diǎn),以保證最佳閱讀效果。本頁提供作業(yè)小助手,一起搜作業(yè)以及作業(yè)好幫手最新版!
    版權(quán)所有 CopyRight © 2012-2024 作業(yè)小助手 All Rights Reserved. 手機(jī)版