求极限lim【1-cosmx)/x^2】,x趋向0
问题描述:
求极限lim【1-cosmx)/x^2】,x趋向0
答
用半角公式
1-cosx=2sin^2(x/2)
所以
(1-cosmx)/x^2
=2sin^2(mx/2)/x^2
然后用等价无穷小
sinx~x,x->0
=2(mx/2)^2/x^2
=m^2/2
极限为m^2/2