[過去ﾛｸﾞ] 高校数学の質問スレ（医者・東大卒専用） Part438 (1002ﾚｽ)
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321: 2024/11/18(月)13:24 ID:EFXQWzbV(1/2) AA×


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321: [sage] 2024/11/18(月) 13:24:25.94 ID:EFXQWzbV Let's approach this step-by-step: Let's make a substitution: t=−1nt=−n1​ As n→∞n→∞, t→0−t→0− (approaching 0 from the negative side) With this substitution, our limit becomes: lim⁡t→0−(1+t)−1tlimt→0−​(1+t)−t1​ Now, we can rewrite this as: lim⁡t→0−((1+t)1t)−1limt→0−​((1+t)t1​)−1 We recognize that lim⁡t→0(1+t)1t=elimt→0​(1+t)t1​=e, which is the definition of Euler's number. This holds true whether t approaches 0 from the positive or negative side. Therefore, our limit becomes: (e)−1=1e(e)−1=e1​ http://rio2016.5ch.net/test/read.cgi/math/1723152147/321

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