高校数学の質問スレ（医者・東大卒専用） Part438

[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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327: １３２人目の素数さん [sage] 2024/11/19(火) 03:57:43.79 ID:V8CHGcRI

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).
Next, we can use the Taylor expansion of ln⁡(1−x)\ln(1 - x)ln(1−x) around x=0x = 0x=0:
ln⁡(1−x)≈−x−x22−x33−…\ln(1 - x) \approx -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldotsln(1−x)≈−x−2x2−3x3−…
For small xxx, we can approximate ln⁡(1−1n)\ln\left(1 - \frac{1}{n}\right)ln(1−n1):
ln⁡(1−1n)≈−1nas n→∞.\ln\left(1 - \frac{1}{n}\right) \approx -\frac{1}{n} \quad \text{as } n \to \infty.ln(1−n1)≈−n1as n→∞.
Thus,
nln⁡(1−1n)≈n(−1n)=−1.n \ln\left(1 - \frac{1}{n}\right) \approx n \left(-\frac{1}{n}\right) = -1.nln(1−n1)≈n(−n1)=−1.
Now, we can take the limit:
lim⁡n→∞nln⁡(1−1n)=−1.\lim_{n \to \infty} n \ln\left(1 - \frac{1}{n}\right) = -1.limn→∞nln(1−n1)=−1.
Exponentiating both sides gives us:
lim⁡n→∞(1−1n)n=e−1=1e.\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = e^{-1} = \frac{1}{e}.limn→∞(1−n1)n=e−1=e1.
Therefore, the final result is:
lim⁡n→∞(1−1n)n=1e.\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.limn→∞(1−n1)n=e1.

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