edited by
8,859 views

8 Answers

Best answer
60 votes
60 votes

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

i.e., -2 constant.

so we get  final ans is $= e^{-2}.$

You can refer this link  for second method

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

Correct Answer: $B$

edited by
4 votes
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

2 votes
2 votes

formula ===> lim x-->∞ ( 1 -   n/x)x = e-n

 

limn---->∞ [   (1 - 1/n) n  ]

limn---->∞ ​​​​​​​[ e-1 ] 2

​​​​​​​[ e-2 ]

Answer:

Related questions

15 votes
15 votes
6 answers
1
makhdoom ghaya asked Oct 2, 2015
3,331 views
The limit of $\dfrac{10^{n}}{n!}$ as $n \to \infty$ is.$0$$1$$e$$10$$\infty$
22 votes
22 votes
4 answers
2
Arjun asked Feb 14, 2017
6,031 views
The value of $\displaystyle \lim_{x\rightarrow 1} \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2}$is $0$is $-1$is $1$does not exist
32 votes
32 votes
6 answers
3
go_editor asked Feb 14, 2015
13,118 views
The value of $\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}$ is$0$$\frac{1}{2}$$1$$\infty$
26 votes
26 votes
6 answers
4
makhdoom ghaya asked Feb 11, 2015
8,154 views
$\displaystyle \lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}}$ is$\infty $$0$$1$Not defined