Line 10: Line 10:
 
==={{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.
+
$$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n} = e^{-2}$$.  
 +
So, option <b>(B)</b> is correct.
  
 
{{Template:FBD}}
 
{{Template:FBD}}

Revision as of 21:35, 14 April 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.




blog comments powered by Disqus



This work is licensed under the CC By-SA 3.0 , without all the cruft that would otherwise be put at the bottom of the page.

Sister Sites: GATE CSE Wiki, GATE CSE, Aptitude Overflow