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

  1. 0
  2. $e^{-2}$
  3. $e^{-1/2}$
  4. 1
asked in Calculus by Veteran (18k points)
edited by | 1.2k views

4 Answers

+24 votes
Best answer

i will solve by two methods 

method 1:

$y =\lim\limits_{n \to \infty}\left(1-\frac{1}{n}\right)^{2n}$

Taking log 

$\log y =\lim\limits_{n \to \infty} 2n \log \left(1-\frac{1}{n}\right)$

$ =\lim\limits_{n \to \infty} \dfrac{\log \left(1-\frac{1}{n}\right)}{\left(\frac{1}{2n}\right)}$---------------(converted this so as to have form $\left(\frac{0}{0}\right)$)

apply L' hospital rule

$\log y=\lim\limits_{n \to \infty} \dfrac{\left(\dfrac{1}{1-\frac{1}{n}}\right).\frac{1}{n^{2}}}{\left(\dfrac{-1}{2n^{2}}\right)}$

$\log y={-2}$

$y=e^{-2}.$


method 2:

it  takes $1$ to power infinity form  

 

$\lim\limits_{x \to \infty} f(x)^{g(x)}$

$=e^{\lim\limits_{x \to \infty} (f(x)-1)g(x)}$

where, $(f(x)-1)*g(x)=\frac{-1}{n}*{2n}={-2}.$

ie -2 constant.

so we get  final ans is $= e^{-2}.$

u can refer this link  for second method

http://www.vitutor.com/calculus/limits/one_infinity.html

answered by Veteran (34.3k points)
edited by
nice answer pooja thanks :)
+4 votes

1infinity form is taken after applying limit 

so general formula for this type of form is elim x->infinty (f(x)-1)*g(x)   here in this question f(x) = (1-(1/n))  and g(x)=2n 

so by aplying above formula elim x->infinity {1-(1/n)-1}*2n =e-2

answered by Boss (5.2k points)
+1 vote

\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{2n}\\ = e^{(-1)*\frac{1}{n}*2n}\\ =e^{-2}

Answer is B.

answered by Loyal (4.2k points)
wrong explanation
0 votes
I think limit of n tends to infinity and answer seems to be 1.

Is it  ?
answered by Boss (9.3k points)
I have updated my question. Now it includes the right answer.

And thank you for pointing out mistake in the question.


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