(cosx-1)/x 的 左极限如何求啊 注意是cosx-1 lim x→0 cosx－1/x²的极限
X\u8d8b\u54110\u65f6\uff0c1-cosx\u4e3a\u4ec0\u4e48\u6ca1\u6709\u5de6\u6781\u9650\uff1f\u6709\u5440\uff0c1-cosx-->1/2 x^2
\u89e3\uff1alim(x\u21920)x²/1-cosx
\u539f\u5f0f=lim(x\u21920)x²/[1-(1-2sin²\uff08x/2\uff09\uff09\u3011
=lim(x\u21920)x²/[2sin²\uff08x/2\uff09\u3011
=lim(x\u21920)x²/\uff08x²/2\uff09
=2
+0=-0=0
当x→0-
两种做法
1 当x足够小时cosx≈1-x^2/2
所以(cosx-1)/x≈-x/2→0+
2 分子分母分别求导数
分子得-sinx→0+
分母得1
所以分式的极限是0+
条件不足，在何处的左极限
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