What is the value of \(\displaystyle\lim_{n \to \infty}\left(1 -…
2010
What is the value of \(\displaystyle\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n}\)
- A.
\(0\) - B.
\(e^{-2}\) - C.
\(e^{-1/2}\) - D.
\(1\)
Attempted by 68 students.
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Correct answer: B
Let L = lim_{n→∞} (1 - 1/n)^{2n}. Take natural logs to simplify: ln L = lim_{n→∞} 2n ln(1 - 1/n).
Rewrite n ln(1 - 1/n) as ln(1 - 1/n) divided by (1/n): n ln(1 - 1/n) = ln(1 - 1/n) / (1/n). As n→∞, (1/n)→0.
Use the standard limit lim_{t→0} ln(1+t)/t = 1 with t = -1/n. This gives ln(1 - 1/n) / (1/n) → -1, so n ln(1 - 1/n) → -1.
Therefore ln L = 2 · (−1) = −2, so L = e^{−2}.
A quicker observation: (1 - 1/n)^{2n} = ((1 - 1/n)^n)^2 and since (1 - 1/n)^n → e^{−1}, the limit is (e^{−1})^2 = e^{−2}.