Question

求极限lim[sinx/(x-pai)] x趋于pai

Answer 1

解法一：原式=lim(x->π)[sin(π-x)/(x-π)] (应用三角函数诱导公式)

=-lim(x->π)[sin(x-π)/(x-π)]

=-1 (应用重要极限lim(x->0)(sinx/x)=1)

解法二：原式=lim(x->π)(cosx) (0/0极限,应用罗比达法则)

=-1