2.1k views

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
in Calculus
edited | 2.1k views

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

Correct Answer: $B$

by Boss (31.4k points)
edited
0
0
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!
0

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

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

by Active (5.1k points)

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

by Active (4.2k points)
–1
wrong explanation
+1 vote

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

limn---->∞ [   (1 - 1/n) n  ]

limn---->∞ ​​​​​​​[ e-1 ] 2

​​​​​​​[ e-2 ]

by Active (4.6k points)
I think limit of n tends to infinity and answer seems to be 1.

Is it  ?
by Loyal (7.2k points)
0
I have updated my question. Now it includes the right answer.

And thank you for pointing out mistake in the question.