方法一：洛必达法则
lim[x→π] sinx/(π-x)
=lim[x→π] cosx/-1
=1
方法二：等价无穷小代换
lim[x→π] sinx/(π-x)
=lim[x→π] sin(π-x)/(π-x)
=lim[x→π] (π-x)/(π-x)
=1