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. $1$

Answer 1

I think limit of n tends to infinity and answer seems to be 1.

Is it  ?

Comments

I have updated my question. Now it includes the right answer.

And thank you for pointing out mistake in the question.

Answer 2

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

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

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 3

Form :- (→1)$^{→\infty}$ => indeterminate form.

$\lim_{n→\infty}$ (1 – 1/n)$^{2n}$
= $\lim_{n→\infty}$ e$^{(ln(1 – 1/n)^{2n})}$
= $\lim_{n→\infty}$ e$^{(2n)ln(1 – 1/n)}$   (let’s find $\lim_{n→\infty}$ n ln(1 – 1/n))

$\lim_{n→\infty}$ n ln(1 – 1/n)  (→$\infty$ x →0  => indeterminate form)
= $\lim_{n→\infty}$ [ln(1 – 1/n)] / (1/n)  (→$\infty$/→$\infty$  => indeterminate form. Apply L’hopital’s rule)
= $\lim_{n→\infty}$ [1/n$^{2}$] / (1 – 1/n)(-1/n$^{2}$) 
= $\lim_{n→\infty}$ (-1)/(1 – 1/n)
= -1

So, $\lim_{n→\infty}$ e$^{(2n)ln(1 – 1/n)}$ = $\lim_{n→\infty}$ e$^{(2)[nln(1 – 1/n)]}$ = $\lim_{n→\infty}$ e$^{(2)(-1)}$ = e$^{-2}$

Another approach :- Use formula : $\lim_{n→a}$ f(n)$^{g(n)}$ where when n→a, then f(n)→1 and g(n)→$\infty$ ((→1)$^{→\infty}$ => indeterminate form)

$\lim_{n→a}$ f(n)$^{g(n)}$ = $\lim_{n→a}$ e$^{g(n)(f(n)–1)}$

$\lim_{n→\infty}$ (1 – 1/n)$^{2n}$ = $\lim_{n→\infty}$ e$^{2n(1 – (1/n) –1)}$ = $\lim_{n→\infty}$ e$^{2n(-1/n)}$ = $\lim_{n→\infty}$ e$^{-2}$ = e$^{-2}$

Answer 4

.