prove that:lim_(n→∞) ∫_(−1) ^1 (1+(t/n))^n dt = e−(1/e).


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Question Number 47098 by rahul 19 last updated on 04/Nov/18

$p r o v e t h a t : \lim_{n \to \infty} \int_{- 1}^{1} {(1 + \frac{t}{n})}^{n} d t = e - \frac{1}{e} .$
Commented by rahul 19 last updated on 04/Nov/18

$M y d o u b t i s w h y w e a r e t a k i n g l i m i t$ $f i r s t a n d t h e n a p p l y i n g i n t e g r a t i o n ? ?$ $I s t h i s c o r r e c t ? ?$
Commented by maxmathsup by imad last updated on 05/Nov/18

$s i r y o u m u s t t a k e a l o o k a n d r e v i s e d o m i n e n t c n v e r g e n c e t h e o r e m b e c a u s e$ $t h e r i s a c o n d i t o n s t o a p p l y t h i s t h e o r e m a n d i t s e a s y t o u s e i t t h a n o t h e r$ $m e t h o d .$
Commented by maxmathsup by imad last updated on 04/Nov/18

$l e t A_{n} = \int_{- 1}^{1} {(1 + \frac{t}{n})}^{n} d t = \int_{R} {(1 + \frac{t}{n})}^{n} χ_{[- 1, 1]} (t) d t l e t f_{n} (t) = {(1 + \frac{t}{n})}^{n} χ_{[- 1, 1]} (t)$ $w e h a v e {l i m}_{n \to + \infty} f_{n} (t) = e^{t} χ_{[- 1, 1]} (t) (s i m p l e c o n v e r g e n c e) a n d$ $∣ f_{n} (t) ∣ ⩽ e^{t} \forall t \in [- 1, 1] d o m i n e n t c o n v e r g e n c e g i v e$ ${l i m}_{n \to + \infty} A_{n} = \int_{R} {l i m}_{n \to} f_{n} (t) d t = \int_{R} e^{t} χ_{[- 1, 1]} (t) d t = \int_{- 1}^{1} e^{t} d t = {[e^{t}]}_{- 1}^{1}$ $= e - \frac{1}{e} .$
Commented by maxmathsup by imad last updated on 04/Nov/18
$another method let A_n = ∫_(−1) ^1 (1+(t/n))^n dt changement (t/n)=x give A_n =∫_(−(1/n)) ^(1/n) (1+x)^n ndx =n ∫_(−(1/n)) ^(1/n) (1+x)^n dx =(n/(n+1)) [(1+x)^(n+1) ]_(−(1/n)) ^(1/n) =(n/(n+1)){(1+(1/n))^(n+1) −(1−(1/n))^(n+1) } but (1+(1/n))^(n+1) =e^((n+1)ln( 1+(1/n))) but we have ln(1+(1/n))=(1/n)+o((1/n)) ⇒ (n+1)ln(1+(1/n))=((n+1)/n) +o(1) →1 (n→+∞) also (n+1)ln(1−(1/n)) =−((n+1)/n) →−1 (n→+∞) ⇒lim_(n→+∞) A_n =lim_(n→+∞) (n/(n+1))(e−e^(−1) ) ⇒ lim_(n→+∞) A_n =e −e^(−1) .$
$a n o t h e r m e t h o d l e t A_{n} = \int_{- 1}^{1} {(1 + \frac{t}{n})}^{n} d t c h a n g e m e n t \frac{t}{n} = x g i v e$ $A_{n} = \int_{- \frac{1}{n}}^{\frac{1}{n}} {(1 + x)}^{n} n d x = n \int_{- \frac{1}{n}}^{\frac{1}{n}} {(1 + x)}^{n} d x$ $= \frac{n}{n + 1} {[{(1 + x)}^{n + 1}]}_{- \frac{1}{n}}^{\frac{1}{n}} = \frac{n}{n + 1} {{(1 + \frac{1}{n})}^{n + 1} - {(1 - \frac{1}{n})}^{n + 1}} b u t$ ${(1 + \frac{1}{n})}^{n + 1} = e^{(n + 1) l n (1 + \frac{1}{n})} b u t w e h a v e l n (1 + \frac{1}{n}) = \frac{1}{n} + o (\frac{1}{n}) \Rightarrow$ $(n + 1) l n (1 + \frac{1}{n}) = \frac{n + 1}{n} + o (1) \to 1 (n \to + \infty) a l s o$ $(n + 1) l n (1 - \frac{1}{n}) = - \frac{n + 1}{n} \to - 1 (n \to + \infty) \Rightarrow {l i m}_{n \to + \infty} A_{n} = {l i m}_{n \to + \infty} \frac{n}{n + 1} (e - e^{- 1}) \Rightarrow$ ${l i m}_{n \to + \infty} A_{n} = e - e^{- 1} .$
Commented by peter frank last updated on 04/Nov/18

$pls help QN 46827$
Commented by rahul 19 last updated on 05/Nov/18

$P r o f A b d o,$ $p l s c l e a r m y d o u b t :$ $I n t h i s p r o b l e m w e a r e f i r s t g i v e n t o$ $i n t e g r a t e a n d t h e n t o a p p l y l i m i t s . . .$ $B u t i f w e f i r s t a p p l y l i m i t s a n d t h e n$ $i n t e g r a t e w e a r e g e t t i n g s a m e r e s u l t$ $a n d w h i c h i s s h o r t e r m e t h o d t o o!$ $S o, i s t h i s m e t h o d c o r r e c t o r i s i t j u s t a$ $c o i n c i d e n c e ?$
Commented by rahul 19 last updated on 05/Nov/18

$o k a y t h a n k y o u p r o f!$ $I^{'} m n e w t o d o m i n e n t c o n v e r g e n c e!!$
Commented by maxmathsup by imad last updated on 06/Nov/18

$n e v e r m i n d s i r m a t h s i s l i k e a s e a w t h o u t b o r d e r s . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

	$I = \int {(1 + \frac{t}{n})}^{n} d t$ $k = 1 + \frac{t}{n} d k = \frac{d t}{n}$ $\int k^{n} \times n d k$ $n \times \frac{k^{n + 1}}{n + 1} + c_{1}$ $s o t h e i n t r e g s l i s$ $\frac{n}{n + 1} \times ∣ {(1 + \frac{t}{n})}^{n + 1} ∣_{- 1}^{1}$ $\frac{n}{n + 1} [{(1 + \frac{1}{n})}^{n + 1} - {(1 - \frac{1}{n})}^{n + 1}$ $n o w a p p l y l i m i t$ $\lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} [{(1 + \frac{1}{n})}^{n + 1} - {(1 - \frac{1}{n})}^{n + 1}]$ $= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \times [\lim_{n \to \infty} (1 + \frac{1}{n}) \times \lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} - \lim_{n \to \infty [} (1 - \frac{1}{n}) \times \lim_{n \to \infty} {(1 - \frac{1}{n})}^{n}]$ $= \frac{1}{1 + 0} \times [(1 + 0) \times e - (1 - 0) \times e^{- 1}]$ $= (e - \frac{1}{e})$ $f i r s t i n t r e g a t i o n t h e n l i m i t d o n e$ $p l s c h e c k . . .$ $n o w f o r m u l a$ $\underset{n \to \infty}{li p}$ $= \underset{n \to \infty}{li l} \times \underset{n \to \infty}{li}$ $n z = \underset{n \to \infty}{li}$ $= \underset{y \to 0}{li}$ $z = e$ $\underset{n \to \infty}{li}$ $l n p = \underset{n \to \infty}{li}$ $w = \frac{1}{n} \underset{w \to 0}{li}$ $= \underset{w \to 0}{li}$
	Commented by rahul 19 last updated on 05/Nov/18
	thanks sir ��
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