题目

x*2-3x+2/sin(x-1)在x=1处的极限

提问时间：2021-12-26

答案

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