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Question Number 186066 by TUN last updated on 31/Jan/23

lim_(x→∞)  x((√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1)  lim_(x→∞)  ((√(x^2 +3x))−((x^3 +2x^2 ))^(1/3) )

limxx(4x212x+7x24x2x+1)limx(x2+3xx3+2x23)

Answered by a.lgnaoui last updated on 31/Jan/23

 x((√(4x^2 −12x+7)) −(√(x^2 −4x−2))  −x+1)  =2x(√(x^2 −3x+(7/4))) −x(√(x^2 −4x−2))  −x^2 +x  lim_(x→∞) [2x(√(x^2 −3x+(7/4))) −x(√(x^2 −4x−2)) −(x^2 −x)]  =lim_(x→∞) [x(((2(4x−3x+((15)/4)))/( (√(x^2 −3x+(7/4))) +(√(x^2 −4x−2)) )))−x(x−1)]  [(((2x^2 )/(x(√(1−(3/x)+(7/(4x^2 )))) +x(√(1−(4/x)−(2/x^2 ))))))−x^2 (1−(1/x))]  =lim_(x→∞) x−x^2 (1−(1/x))=lim_(x→∞) (−x^2 )=−∞

x(4x212x+7x24x2x+1)=2xx23x+74xx24x2x2+xlimx[2xx23x+74xx24x2(x2x)]=limx[x(2(4x3x+154)x23x+74+x24x2)x(x1)][(2x2x13x+74x2+x14x2x2)x2(11x)]=limxxx2(11x)=limx(x2)=

Commented by Frix last updated on 01/Feb/23

Wrong.

Wrong.

Answered by Frix last updated on 01/Feb/23

lim_(x→∞)  x((√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1) =  lim_(t→0^+ )  ((t−1+(√(7t^2 −12t+4))−(√(−2t^2 +4t+1)))/t^2 ) =  =lim_(t→0^+ )  (((d^2 [t−1+(√(7t^2 −12t+4))−(√(−2t^2 +4t+1))])/dt^2 )/((d^2 [t^2 ])/dt^2 )) =  =lim_(t→0^+ )  (((6/((−2t^2 −4t+1)^(3/2) ))−(8/((7t^2 −12t+4)^(3/2) )))/2) =  =(5/2)  Similar procedure for the 2^(nd)  one leads to (5/6)

limxx(4x212x+7x24x2x+1)=limt0+t1+7t212t+42t2+4t+1t2==limt0+d2[t1+7t212t+42t2+4t+1]dt2d2[t2]dt2==limt0+6(2t24t+1)328(7t212t+4)322==52Similarprocedureforthe2ndoneleadsto56

Answered by cortano1 last updated on 01/Feb/23

Without L′Hopital    L=lim_(x→∞) x[(√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1 ]  = lim_(x→∞) x[ (√(4x^2 −12x+7))−(2x−3)+(x−2)−(√(x^2 −4x−2)) ]  = lim_(x→∞) x[((−2)/( (√(4x^2 −12x+7))+(2x−3))) + (6/(x−2+(√(x^2 −4x−2)))) ]  = ((−2)/( (√4)+2)) +(6/(1+1)) = (5/2)

WithoutLHopitalL=limxx[4x212x+7x24x2x+1]=limxx[4x212x+7(2x3)+(x2)x24x2]=limxx[24x212x+7+(2x3)+6x2+x24x2]=24+2+61+1=52

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