x*2-3x+2/sin(x-1)在x=1处的极限

2个回答

  • 解法一:原式=lim(x->1)[(x-1)(x-2)/sin(x-1)] (把分式分子分解因式)

    =lim(x->1){[(x-1)/sin(x-1)]*(x-2)}

    =lim(x->1)[(x-1)/sin(x-1)]*lim(x->1)(x-2)

    =1*(-1) ()应用重要极限lim(x->0)(sinx/x)=1)

    =-1

    解法二:原式=lim(x->1)[(2x-3)/cos(x-1)] (0/0型极限,应用罗比达法则)

    =(2*1-3)/cos(1-1)

    =-1/1

    =-1