lim[(1-cosx)^1/2]/sinx,x趋于0,求极限

用等价无穷小替换
原式=lim(x→0)√(2sin^2(x/2))/sinx
=lim(x→0)√2|sin(x/2)|/sinx
因为右极限为lim(x→0+)√2*sin(x/2)/sinx=lim(x→0)√2*(x/2)/2=√2/2
类似地,左极限为-√2/2
所以极限不存在.就是右极限啦。右极限为lim(x→0+)√2*sin(x/2)/sinx=lim(x→0)√2*(x/2)/2=√2/2