洛必达法则例题

2025-05-09 06:10:25
推荐回答(1个)
回答1:

1-cosx = 1-{1-2[sin(x/2)]^2} = 2[sin(x/2)]^2
xsinx = 2xsin(x/2)cos(x/2)
原式= lim 2[sin(x/2)]^2 / [2xsin(x/2)cos(x/2)] = tgx / x
对分子分母同时求导(洛必达法则)
(tgx)' = 1 / (cosx)^2
(x)' = 1
原式 = lim 1/(cosx)^2
当 x --> 0 时,cosx ---> 1
原式 = 1