GATE CSE 2010 | Question: 5

Question

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

• A. $0$
• B. $e^{-2}$
• C. $e^{-1/2}$
• D. $1$

Comments

why can't we use binomial theorem in this?

Answer 1

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

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

$\log y={-2}$

$y=e^{-2}.$

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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: $B$

Comments

nice answer pooja thanks :)

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What's wrong if I substitute value of n which is infinity. Doing so I get answer as 1, so why is it going wrong? Please anyone help!

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@ShivamB 

$1^{\infty }$ form is one of the indeterminate-forms because taking its log results into $\infty \cdot 0$ form.

Answer 2

1^infinity form is taken after applying limit 

so general formula for this type of form is e^{lim 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 e^{lim x->infinity {1-(1/n)-1}*2n} =e^-2

Answer 3

formula ===> lim _{x-->∞ ( 1 -   n/x)^x = e^-n}

 

lim_n---->∞ [   (1 - 1/n) ⁿ  ]² 

lim_n---->∞ ​​​​​​​[ e^-1 ] ²

​​​​​​​[ e^-2 ]

Answer 4

Answer is B.

Comments

wrong explanation

Answer 5

$y=\lim_{n->\infty}(1-\frac{1}{n})^{2n}$

$\log y=\lim_{n->\infty}2n*\log(1-\frac{1}{n})$

let , $z=-\frac{1}{n}$

as $n->\infty$  , $z->0$

$\log y=(-2)\lim_{z->0}\frac{\log (1+z)}{z}$

as we know $\lim_{z->0}\frac{\log (1+z)}{z}=1$

$\log y=(-2)*1$

$y=e^{-2}$

correct answer (B)

Answer 6

.

Comments

Small correction. Lim x-> inf and not n -> inf at the formula