高等数学第十题
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(10) lim<x→1>[x/(x-1) - 1/lnx] = lim<x→1>(xlnx-x+1)/[(x-1)lnx] (令 x = 1+u)
= lim<u→0>[(1+u)ln(1+u)-u]/[uln(1+u)] = lim<u→0>[(1+u)ln(1+u)-u]/u^2 (0/0)
= lim<u→0>[ln(1+u)+1-1]/(2u) = lim<u→0> u/(2u) = 1/2
= lim<u→0>[(1+u)ln(1+u)-u]/[uln(1+u)] = lim<u→0>[(1+u)ln(1+u)-u]/u^2 (0/0)
= lim<u→0>[ln(1+u)+1-1]/(2u) = lim<u→0> u/(2u) = 1/2
追问
为什么前面那步会等于ln(1+u)+1-1?
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