求极限x趋向于0 (1-cos2x)⼀xsinx

2025-01-19 20:38:53
推荐回答(3个)
回答1:

用等价无穷小做:当x→0时 1-cos(2x)~(1/2)x²
sin(x)~x
所以 lim(x→0)(1-cos2x)/xsinx
=lim(x→0) (x^2/2)/x^2
=1/2

回答2:

lim(x→0)(1-cos2x)/xsinx
=lim(x→0) (x^2/2)/x^2
=1/2

回答3:

=lim{1-[1-2(sinx)^2]}/xsinx
=lim2sinx/x
=2