(Created page with "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 ==={{Template:Author|Happy Mitta...")
 
 
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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  
$$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$. So option <b>(B)</b> is correct.
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$$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$
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So, option <b>(B)</b> is correct.
  
 
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[[Category:Calculus]]
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[[Category: GATE2010]]
 
[[Category: GATE2010]]
[[Category: Previous year GATE questions]]
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[[Category: Calculus questions from GATE]]

Latest revision as of 11:50, 15 July 2014

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.




blog comments powered by Disqus