Question

求极限lim n→0 (根号下1-cosx)/sinx

Answer 1

1-cosx=1-(1-2sin²x/2)=2sin²x/2

所以

x→0-

原式=lim-√2*sin(x/2)/(2sin(x/2)cos(x/2))

=lim-√2/(2cos(x/2))

=-√2/2

x→0+

原式=lim√2*sin(x/2)/(2sin(x/2)cos(x/2))

=lim√2/(2cos(x/2))

=√2/2

不相等

所以极限不存在