limx->π/2 cosx/(x-π/2)

limx->π/2 cosx/(x-π/2)
limx->π/2 cosx/(x-π/2)

limx->π/2 cosx/(x-π/2)
令a=π/2-x
则x=π/2-a
所以原式=lim(a趋于0)cos(π/2-a)/(-a)
=-lim(a趋于0)sina/a
=-1

洛必达
上下同求导
=-sinx/1|x=π/2
=-sinπ/2
=-1

limx->π/2 cosx/(x-π/2)=-limx->π/2 sin(π/2-x)/(π/2-x)=-1

分子的导：cosx‘=-sinx分母的导：(x-π/2)’=1因为X->π/2 -sinx=-1相除=-1

应该利用的是那个什么定理，就是分子分母都求导之后极限还不变的那个。（哦~原来是洛必达~）
lim(x→π/2) cosx/(x-π/2)=lim(x→π/2) -sinx/1 =-1