一道极限题lim(x→0)√(tansinx)/(sintanx)

4个回答

  • 先不管根号

    就看

    tansinx/sintanx

    当x->0时 是 0/0的形式

    所以应用罗必塔法则,即对分子分母分别求导得

    (tansinx)'=1/cos^2(sinx) * cosx=cosx/cos^2(sinx)

    (sintanx)'=costanx*1/cos^2x=costanx/cos^2x

    所以

    (tansinx)'/(sintanx)'=cosx/cos^2(sinx) / costanx/cos^2x = cos^3x / costanx * cos^2(sinx)

    所以

    lim(x->0)√(tansinx)/(sintanx)=lim(x->0)√cos^3x / costanx * cos^2(sinx)

    =√1/cos(0)*cos^2(0)

    =√1/1

    =1