Question Number 148655 by mathdanisur last updated on 29/Jul/21

lim_(n→∞) ∫_( 0) ^( 1) ((nx)/(1 + n^2 x^4 )) dx = ?

Answered by ArielVyny last updated on 29/Jul/21

n∫_0 ^1 (x/(1+(nx^2 )^2 ))dx=_(nx^2 =t) n∫_0 ^n ((√(t/n))/(1+t^2 )) n2xdx=dt→dx=(1/(2nx))dt=(1/(2n(√(t/n))))dt n∫_0 ^n ((√(t/n))/(1+t^2 ))×(1/(2n(√(t/n))))dt=(1/2)∫_0 ^n (1/(1+t^2 ))dt=(1/2)arctg(n) conclusion lim_(n→∞) ∫_0 ^1 ((nx)/(1+n^2 x^4 ))dx=(π/4)

Commented bymathdanisur last updated on 29/Jul/21

Thankyou Ser No solution or (π/4) .?

Commented byArielVyny last updated on 29/Jul/21

if you have any solutions you can post sir but i think the solution is (π/4)

Commented bymathdanisur last updated on 29/Jul/21

Thanks Ser

Answered by Ar Brandon last updated on 30/Jul/21

L=lim_(n→∞) ∫_0 ^1 ((nx)/(1+n^2 x^4 ))dx=(1/2)lim_(n→∞) ∫_0 ^1 ((2nx)/(1+(nx^2 )^2 ))dx =(1/2)lim_(n→∞) [arctan(nx^2 )]_0 ^1 =(1/2)[lim_(n→∞) arctan(n)−lim_(n→∞, x→0) arctan(nx^2 )] =(π/4)−(1/2)lim_(n→∞, x→0) arctan(nx^2 ) No solution L oscillates within 0≤L≤(π/4)

Commented bymathdanisur last updated on 29/Jul/21

Thank you Ser

Commented bymathmax by abdo last updated on 30/Jul/21

not correct your way sir..!

Commented byAr Brandon last updated on 02/Aug/21

What about lim_(x→0, n→∞) arctan(nx), Sir ?

Answered by mathmax by abdo last updated on 30/Jul/21

A_n =∫_0 ^1 ((nx)/(1+n^2 x^4 ))dx changement nx^2 =t give 2nxdx=dt ⇒ A_n =∫_0 ^n (dt/(2(1+t^2 ))) =(1/2)[arctant]_0 ^n =(1/2)arctan(n) ⇒ lim_(n→+∞) A_n =(1/2).(π/2)=(π/4)

Commented bymathdanisur last updated on 31/Jul/21

Thank You Ser but answer: no solution

Commented bymathmax by abdo last updated on 31/Jul/21

the answer is (π/4)sir

Commented bymathdanisur last updated on 01/Aug/21

Thankyou Sir