cosx-1/x 极限的详细解法?
题目是cosx-1/x而不是(cosx-1)/x吗?那这样就简单了
x趋于0
lim.x->0[cosx-1/x]=lim.x->0[1-2(sinx/2)^2-1/x]=1
绛旓細cosx-1=cos(x/2+x/2)-1 =[cos(x/2)]^2-[sin(x/2)]^2-([cos(x/2)]^2+[sin(x/2)]^2)=-2sin(x/2)^2
绛旓細cosx-1=-2sin(x/2)^2銆俢osx-1=cos(x/2+x/2)-1=[cos(x/2)]^2-[sin(x/2)]^2-([cos(x/2)]^2+[sin(x/2)]^2)=-2sin(x/2)^2涓夎鍑芥暟鐨勫畾涔夛細涓夎鍑芥暟鏄熀鏈垵绛夊嚱鏁颁箣涓锛屾槸浠ヨ搴︼紙鏁板涓婃渶甯哥敤寮у害鍒讹紝涓嬪悓锛変负鑷彉閲忥紝瑙掑害瀵瑰簲浠绘剰瑙掔粓杈逛笌鍗曚綅鍦嗕氦鐐瑰潗鏍囨垨鍏舵瘮鍊间负鍥犲彉...
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