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

Revision as of 21:36, 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

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