2.5计算极限lim(x→0) (1-cos2x)/xsinx
人气：325 ℃ 时间：2020-05-06 23:08:08
解答
cos2x=1-2sin²x
(1-cos2x)/xsinx=[1-((1-2sin²x)]/xsinx
=2sin²x/xsinx
=2sinx/x
lim(x→0) (1-cos2x)/xsinx=lim(x→0) 2sinx/x=2
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