Question

求lim(x→0)[(x-sinx)/(x+sinx)],不能用洛必达法则，并且只能使用函数极限与连续的知识解答

请用第一章函数极限与连续的知识解答

Answer 1

lim(x->0) [(x-sinx)/(x+sinx)]

=lim(x->0) (x/sinx -1)/(x/sinx+1)
= (1-1)/(1+1)
=0

追问

0

追答

lim(x->0) [(x-sinx)/(x+sinx)]

=lim(x->0) (x/sinx -1)/(x/sinx+1)
= (1-1)/(1+1)
=0

追问

为什么分子分母同除以sinx，我觉得应该同除以x，再利用lim（x→0）sinx/x.

追答

lim(x->0) sinx/x = 1
lim(x->0) x/sinx = 1

那是一样
lim(x->0) [(x-sinx)/(x+sinx)]

=lim(x->0) (1 -sinx/x)/(1+ sinx/x)
= (1-1)/(1+1)
=0