[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) AAS
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​

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