高校数学の質問スレ（医者・東大卒専用） Part438 (979ﾚｽ)
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327: 2024/11/19(火)03:57 ID:V8CHGcRI(1/2) AAS
To evaluate the limit
lim⁡n→∞(1−1n)n,\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n,limn→∞​(1−n1​)n,
we can recognize that this expression is related to the definition of the number eee. Specifically, we can rewrite the expression in a more convenient form.
First, we can use the fact that
(1−1n)n=((1−1n)−n)−1.\left(1 - \frac{1}{n}\right)^n = \left(\left(1 - \frac{1}{n}\right)^{-n}\right)^{-1}.(1−n1​)n=((1−n1​)−n)−1.
Now, we can take the natural logarithm of the expression to simplify the limit:
ln⁡((1−1n)n)=nln⁡(1−1n).\ln\left(\left(1 - \frac{1}{n}\right)^n\right) = n \ln\left(1 - \frac{1}{n}\right).ln((1−n1​)n)=nln(1−n1​).
省12

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328: 2024/11/19(火)05:02 ID:V8CHGcRI(2/2) AAS
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