原式lim(x→1) sin(x-1)/(x^2-1)
根据洛必达法则，lim f(x)/g(x) = limf"(x)/g"(x)
所以lim(x→1) sin(x-1)/(x^2-1) ====》limcos(x-1)/2x = 1/2
分子求导/分母求导
求极限 lim(x→1) sin(x-1)/(x^2-1)
解：
lim(x→1)sin(x-1)/(x²-1)
=lim(x→1)sin(x-1)/(x+1)(x-1)
=lim(x→1)[1/(x+1)*sin(x-1)/(x-1)]
=(1/2)*1
=1/2
令a=x-1
则a→0
且x²-1=(a+1)²-1=a(a+2)
所以原式=lim(a→0)sina/a(a+2)
=lim(a→0)sina/a*lim(a→0)1/(a+2)
=1*1/(0+2)
=1/2
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