2、求极限 lim┬(x→π/2) ( π/2-x) tanx .

问题描述:

2、求极限 lim┬(x→π/2) ( π/2-x) tanx .

0
(π/2-x)*sin(π/2-x)/cos(π/2-x)
0*0/1=0

令t=π/2-x,
x=π/2-t,
则lim┬(t→0)t tan(π/2-t)=lim┬(t→0)t cot(t)=1(重要极限)

lim┬(x→π/2) ( π/2-x) tanx =lim┬(x→π/2) ( π/2-x) sinx/cosx
=lim┬(x→π/2) [( π/2-x)/ sin(π/2-x)]*sinx
=lim┬(x→π/2) [( π/2-x)/ sin(π/2-x)]*lim┬(x→π/2)sinx
=1.
(注释:sin(π/2-x)=cosx)