当t接近tanδt?/ tan5t的0时,限制是多少

当t接近tanδt?/ tan5t的0时,限制是多少
Anonim

回答:

#Lt(叔> 0)(tan8t)/(tan5t)= 8/5#

说明:

让我们先找到 #Lt_(X-> 0)坦/ X#

#Lt_(X-> 0)坦/ X = Lt_(X-> 0)(sinx的)/(xcosx)#

= #Lt_(x-> 0)(sinx)/ x xx Lt_(x-> 0)1 / cosx#

= #1xx1 = 1#

于是 #Lt_(叔> 0)(tan8t)/(tan5t)#

= #Lt_(叔> 0)((tan8t)/(8T))/((tan5t)/(5T))XX(8T)/(5T)#

= #(Lt_(8t-> 0)((tan8t)/(8T)))/(Lt_(5t-> 0)((tan5t)/(5T)))XX8 / 5#

= #1 / 1xx8 / 5 = 8/5#