计算极限lim(x-0)，(1-cos2x)/xsin2x 求limx→0(1sin2x?cos2xx2)
\u6c42\u6781\u9650limx 0\u65f61-cos2x/sinx\u7b49\u4e8e\u591a\u5c11
\u89c1\u56fe
\u2235x\u21920\u65f6\uff0csin2x\uff5ex2\uff0c1-cos4x\uff5e(4x)22\uff1d8x2\u2234limx\u21920(1sin2x?cos2xx2)\uff1dlimx\u21920x2?sin2xcos2xx2sin2x=limx\u21920x2?14sin22xx4\uff1dlimx\u219202x?12sin4x4x3\uff1dlimx\u219201?cos4x6x2\uff1dlimx\u2192012(4x)26x2\uff1d43\uff0e
利用等价无穷小，当x->0时，1-cos2x等价于(2x)^2/2=2x^2而sin2x等价于2x
所以，原极限值=2x^2/(x*2x)=1
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