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Question Number 47098 by rahul 19 last updated on 04/Nov/18
prove that:lim_(n→∞)  ∫_(−1) ^1 (1+(t/n))^n dt = e−(1/e).
$${prove}\:{that}:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} {dt}\:=\:{e}−\frac{\mathrm{1}}{{e}}. \\ $$
Commented by rahul 19 last updated on 04/Nov/18
My doubt is why we are taking limit  first and then applying integration??  Is this correct??
$${My}\:{doubt}\:{is}\:{why}\:{we}\:{are}\:{taking}\:{limit} \\ $$$${first}\:{and}\:{then}\:{applying}\:{integration}?? \\ $$$${Is}\:{this}\:{correct}?? \\ $$
Commented by maxmathsup by imad last updated on 05/Nov/18
sir you must take a look  and revise dominent cnvergence theorem because   ther is a conditons to apply this theorem and its easy to use it than other  method.
$${sir}\:{you}\:{must}\:{take}\:{a}\:{look}\:\:{and}\:{revise}\:{dominent}\:{cnvergence}\:{theorem}\:{because}\: \\ $$$${ther}\:{is}\:{a}\:{conditons}\:{to}\:{apply}\:{this}\:{theorem}\:{and}\:{its}\:{easy}\:{to}\:{use}\:{it}\:{than}\:{other} \\ $$$${method}. \\ $$
Commented by maxmathsup by imad last updated on 04/Nov/18
let A_n = ∫_(−1) ^1 (1+(t/n))^n dt =∫_R (1+(t/n))^n χ_([−1,1]) (t)dt let f_n (t)=(1+(t/n))^n  χ_([−1,1]) (t)  we have lim_(n→+∞) f_n (t) =e^t χ_([−1,1]) (t) (simple convergence ) and  ∣f_n (t)∣ ≤e^t   ∀ t ∈[−1,1] dominent convergence give  lim_(n→+∞) A_n =∫_R lim_(n→)  f_n (t)dt = ∫_R e^t  χ_([−1,1]) (t)dt =∫_(−1) ^1  e^t dt=[e^t ]_(−1) ^1   =e −(1/e) .
$${let}\:{A}_{{n}} =\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} {dt}\:=\int_{{R}} \left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} \chi_{\left[−\mathrm{1},\mathrm{1}\right]} \left({t}\right){dt}\:{let}\:{f}_{{n}} \left({t}\right)=\left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} \:\chi_{\left[−\mathrm{1},\mathrm{1}\right]} \left({t}\right) \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} {f}_{{n}} \left({t}\right)\:={e}^{{t}} \chi_{\left[−\mathrm{1},\mathrm{1}\right]} \left({t}\right)\:\left({simple}\:{convergence}\:\right)\:{and} \\ $$$$\mid{f}_{{n}} \left({t}\right)\mid\:\leqslant{e}^{{t}} \:\:\forall\:{t}\:\in\left[−\mathrm{1},\mathrm{1}\right]\:{dominent}\:{convergence}\:{give} \\ $$$${lim}_{{n}\rightarrow+\infty} {A}_{{n}} =\int_{{R}} {lim}_{{n}\rightarrow} \:{f}_{{n}} \left({t}\right){dt}\:=\:\int_{{R}} {e}^{{t}} \:\chi_{\left[−\mathrm{1},\mathrm{1}\right]} \left({t}\right){dt}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:{e}^{{t}} {dt}=\left[{e}^{{t}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$={e}\:−\frac{\mathrm{1}}{{e}}\:. \\ $$
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)  .
$${another}\:{method}\:\:{let}\:{A}_{{n}} =\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} \:{dt}\:\:{changement}\:\frac{{t}}{{n}}={x}\:{give} \\ $$$${A}_{{n}} =\int_{−\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{1}}{{n}}} \:\left(\mathrm{1}+{x}\right)^{{n}} \:{ndx}\:={n}\:\int_{−\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{1}}{{n}}} \:\left(\mathrm{1}+{x}\right)^{{n}} {dx} \\ $$$$=\frac{{n}}{{n}+\mathrm{1}}\:\left[\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} \right]_{−\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{1}}{{n}}} \:=\frac{{n}}{{n}+\mathrm{1}}\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \right\}\:{but} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \:={e}^{\left({n}+\mathrm{1}\right){ln}\left(\:\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)} \:\:{but}\:{we}\:{have}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$$\left({n}+\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=\frac{{n}+\mathrm{1}}{{n}}\:+{o}\left(\mathrm{1}\right)\:\rightarrow\mathrm{1}\:\left({n}\rightarrow+\infty\right)\:{also} \\ $$$$\left({n}+\mathrm{1}\right){ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\:=−\frac{{n}+\mathrm{1}}{{n}}\:\rightarrow−\mathrm{1}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \frac{{n}}{{n}+\mathrm{1}}\left({e}−{e}^{−\mathrm{1}} \right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={e}\:−{e}^{−\mathrm{1}} \:. \\ $$$$ \\ $$
Commented by peter frank last updated on 04/Nov/18
pls help QN 46827
$$\mathrm{pls}\:\mathrm{help}\:\mathrm{QN}\:\mathrm{46827} \\ $$
Commented by rahul 19 last updated on 05/Nov/18
Prof Abdo,  pls clear my doubt:  In this problem we are first given to  integrate and then to apply limits...  But if we first apply limits and then  integrate we are getting same result  and which is shorter method too!  So, is this method correct or is it just a   coincidence?
$${Prof}\:{Abdo}, \\ $$$${pls}\:{clear}\:{my}\:{doubt}: \\ $$$${In}\:{this}\:{problem}\:{we}\:{are}\:{first}\:{given}\:{to} \\ $$$${integrate}\:{and}\:{then}\:{to}\:{apply}\:{limits}… \\ $$$${But}\:{if}\:{we}\:{first}\:{apply}\:{limits}\:{and}\:{then} \\ $$$${integrate}\:{we}\:{are}\:{getting}\:{same}\:{result} \\ $$$${and}\:{which}\:{is}\:{shorter}\:{method}\:{too}! \\ $$$${So},\:{is}\:{this}\:{method}\:{correct}\:{or}\:{is}\:{it}\:{just}\:{a}\: \\ $$$${coincidence}? \\ $$
Commented by rahul 19 last updated on 05/Nov/18
okay thank you prof !  I′m new to dominent convergence !!
$${okay}\:{thank}\:{you}\:{prof}\:! \\ $$$${I}'{m}\:{new}\:{to}\:{dominent}\:{convergence}\:!! \\ $$
Commented by maxmathsup by imad last updated on 06/Nov/18
nevermind sir maths is like a sea wthout borders....
$${nevermind}\:{sir}\:{maths}\:{is}\:{like}\:{a}\:{sea}\:{wthout}\:{borders}…. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18
I=∫(1+(t/n))^n dt  k=1+(t/n)   dk=(dt/n)  ∫k^n ×ndk  n×(k^(n+1) /(n+1))+c_1   so the intregsl is  (n/(n+1))×∣(1+(t/n))^(n+1) ∣_(−1) ^1   (n/(n+1))[(1+(1/n))^(n+1) −(1−(1/n))^(n+1)   now apply limit  lim_(n→∞)  (1/(1+(1/n)))[(1+(1/n))^(n+1) −(1−(1/n))^(n+1) ]  =lim_(n→∞) (1/(1+(1/n)))×[lim_(n→∞) (1+(1/n))×lim_(n→∞) (1+(1/n))^n −lim_(n→∞[) (1−(1/n))×lim_(n→∞) (1−(1/n))^n ]  =(1/(1+0))×[(1+0)×e−(1−0)×e^(−1) ]  =(e−(1/e))  first intregation then limit done  pls check...  now formula                                                            lip_(n→∞)   =lil_(n→∞) ×li_(n→∞)   nz=li_(n→∞)   =li_(y→0)   z=e  li_(n→∞)   lnp=li_(n→∞)   w=(1/n)         li_(w→0)   =li_(w→0)
$${I}=\int\left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} {dt} \\ $$$${k}=\mathrm{1}+\frac{{t}}{{n}}\:\:\:{dk}=\frac{{dt}}{{n}} \\ $$$$\int{k}^{{n}} ×{ndk} \\ $$$${n}×\frac{{k}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+{c}_{\mathrm{1}} \\ $$$${so}\:{the}\:{intregsl}\:{is} \\ $$$$\frac{{n}}{{n}+\mathrm{1}}×\mid\left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}+\mathrm{1}} \mid_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$\frac{{n}}{{n}+\mathrm{1}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \right. \\ $$$${now}\:{apply}\:{limit} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}×\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} −\underset{{n}\rightarrow\infty\left[\right.} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)^{{n}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}×\left[\left(\mathrm{1}+\mathrm{0}\right)×{e}−\left(\mathrm{1}−\mathrm{0}\right)×{e}^{−\mathrm{1}} \right] \\ $$$$=\left({e}−\frac{\mathrm{1}}{{e}}\right) \\ $$$${first}\:{intregation}\:{then}\:{limit}\:{done} \\ $$$${pls}\:{check}… \\ $$$${now}\:{formula} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{li}{p}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{li}{l}}×\underset{{n}\rightarrow\infty} {\mathrm{li}} \\ $$$${nz}=\underset{{n}\rightarrow\infty} {\mathrm{li}} \\ $$$$=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$$${z}={e} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{li}} \\ $$$${lnp}=\underset{{n}\rightarrow\infty} {\mathrm{li}} \\ $$$${w}=\frac{\mathrm{1}}{{n}}\:\:\:\:\:\:\:\:\:\underset{{w}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$$$=\underset{{w}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$
Commented by rahul 19 last updated on 05/Nov/18
thanks sir ����

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