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Question Number 129120 by liberty last updated on 13/Jan/21

  lim_((x,y)→(∞,∞)) ((π/2) −arctan (x^2 +y^2 ))^(1/(ln (x^2 +y^2 ))) =?

lim(x,y)(,)(π2arctan(x2+y2))1ln(x2+y2)=?

Answered by benjo_mathlover last updated on 13/Jan/21

L=lim_((x,y)→(∞,∞)) ((π/2)−tan^(−1) (x^2 +y^2 ))^(1/(ln (x^2 +y^2 )))   L=e^(lim_((x,y)→(∞,∞)) ((((π/2)−tan^(−1) (x^2 +y^2 ))/(ln (x^2 +y^2 )))))   L=e^(lim_((x,y)→(∞,∞)) (((−((2x+2y)/(1+(x^2 +y^2 )^2 )))/((2x+2y)/(x^2 +y^2 ))) ))   L=e^(lim_((x,y)→(∞,∞)) −(((x^2 +y^2 )/(1+(x^2 +y^2 )^2 ))))   ((L= e^0  = 1)/)

L=lim(x,y)(,)(π2tan1(x2+y2))1ln(x2+y2)L=elim(x,y)(,)(π2tan1(x2+y2)ln(x2+y2))L=elim(x,y)(,)(2x+2y1+(x2+y2)22x+2yx2+y2)L=elim(x,y)(,)(x2+y21+(x2+y2)2)L=e0=1

Answered by mathmax by abdo last updated on 14/Jan/21

we do the changement x=rcosθ and y=rsinθ ⇒  Lim=lim_(r→+∞) ((π/2)−arctan(r^2 ))^(1/(2lnr))  =lim_(r→∞) e^((1/(2lnr))ln((π/2) −arctan(r^2 )))   we have∣ (π/2)−arctan(r^2 )∣<π ⇒(1/(2lnr))ln((π/2)−arctan(r^2 ))<(π/(2lnr))→o(r→+∞) ⇒  lim_(r→0)  e^((1/(2nr))ln((π/2)−arctan(r^2 ))) =e^0  =1

wedothechangementx=rcosθandy=rsinθLim=limr+(π2arctan(r2))12lnr=limre12lnrln(π2arctan(r2))wehaveπ2arctan(r2)∣<π12lnrln(π2arctan(r2))<π2lnro(r+)limr0e12nrln(π2arctan(r2))=e0=1

Commented by mathmax by abdo last updated on 14/Jan/21

0<θ<(π/2)

0<θ<π2

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