已知x不等于1时，f（x）=lncos（x-1）/1-sin派x/2，试定义f（1）的值，使f(x lim(x→1) lncos(x-1)/(1-sin(πx/...
lim(x\u21921) lncos(x-1)/(1-sin(\u03c0x/2))\u539f\u5f0f=lim(x\u21921) ln[1+cos(x-1)-1]/(1-sin(\u03c0x/2))
=lim(x\u21921) [cos(x-1)-1]/(1-sin(\u03c0x/2))
=lim(x\u21921)[-1/2(x-1)^2]/(1-sin(nx/2)
\u63a5\u4e0b\u5c31\u5b8c\u5168\u662f\u5229\u7528\u6d1b\u6bd4\u8fbe\u6cd5\u5219\u4e86\uff0c\u9274\u4e8e\u7b26\u53f7\u592a\u96be\u6253\u4e86\uff0c\u5c31\u4e0d\u8d58\u8ff0\u4e86
\u7b80\u5355\u8ba1\u7b97\u4e00\u4e0b\u5373\u53ef\uff0c\u7b54\u6848\u5982\u56fe\u6240\u793a
简单计算一下即可，答案如图所示
x→1 时，分子 lncos(x - 1)
＝ln[1＋cos(x - 1) - 1]
≈ cos(x-1) - 1
≈ - (x-1)² / 2，
分母 1 - sin(πx/2)
＝1 - cos(πx/2 - π/2)
＝1 - cos[π/2 * (x-1)]
≈ (π/2)²(x-1)² / 2，
所以极限＝ - 4/π²，
定义 f(1)＝ - 4/π² 。
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