Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
We know that $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = e^{-1}$, so | We know that $\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = e^{-1}$, so | ||
| − | $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$ | + | $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$ |
So, option <b>(B)</b> is correct. | So, option <b>(B)</b> is correct. | ||
{{Template:FBD}} | {{Template:FBD}} | ||
| − | + | ||
[[Category: GATE2010]] | [[Category: GATE2010]] | ||
| − | [[Category: | + | [[Category: Calculus questions from GATE]] |
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
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.
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
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.
GATECSE