What is the value of $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n}$ ?

(A) 0

(B) $e^{-2}$

(C) $e^{-1/2}$

(D) 1

Solution by Happy Mittal

We know that $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = e^{-1}$, so $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$. So, option (B) is correct.




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What is the value of $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n}$ ?

(A) 0

(B) $e^{-2}$

(C) $e^{-1/2}$

(D) 1

Solution by Happy Mittal[edit]

We know that $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = e^{-1}$, so $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$. So, option (B) is correct.




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