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lim-x-8x-3-4x-2-1-1-3-8x-3-px-2-1-1-3-1-4-find-p-

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Question Number 82175 by jagoll last updated on 19/Feb/20
lim_(x→−∞) ((8x^3 −4x^2 +1))^(1/(3 )) −((8x^3 +∣px∣^2 −1))^(1/(3 )) = (1/4) find p
limx→−∞8x3−4x2+13−8x3+∣px∣2−13=14findp
Commented by mr W last updated on 19/Feb/20
please check sir! question is wrong. the limit is ≤−(1/3). it can neither be (1/4) nor −(1/4).
pleasechecksir!questioniswrong.thelimitis⩽−13.itcanneitherbe14nor−14.
Commented by john santu last updated on 19/Feb/20
lim_(u→0^− ) ((((8−4u+u^3 ))^(1/(3 )) − ((8+∣p∣ u−u^3 ))^(1/(3 )) )/u) = lim_(u→0^− ) ((((3u^2 −4)/(3((((8−4u+u^3 )^2 ))^(1/(3 )) ))− (((−3u^2 +∣p∣))/(3((((8+∣p∣u−u^3 )^2 ))^(1/(3 )) )))/1) = ((−4)/(3.4))−((∣p∣)/(3.4)) = (1/4) ⇒ −(1/3)−1= ((∣p∣)/3) the question is wrong
limu→0−8−4u+u33−8+∣p∣u−u33u=limu→0−3u2−43((8−4u+u3)23−(−3u2+∣p∣)3((8+∣p∣u−u3)231=−43.4−∣p∣3.4=14⇒−13−1=∣p∣3thequestioniswrong
Commented by mathmax by abdo last updated on 19/Feb/20
we have^3 (√(8x^3 −4x^2 +1))=^3 (√(8x^3 (1−(1/(2x))+(1/(8x^3 ))))) =2x(√(1−(1/(2x))+(1/(8x^3 )))) ∼ 2x(1+(1/2)(−(1/(2x)) +(1/(8x^3 )))) (x→−∞) =2x{1−(1/(4x)) +(1/(16x^3 ))} =2x−(1/2) +(1/(8x^2 )) also we have^3 (√(8x^3 +∣p∣x^2 −1))=^3 (√(8x^3 (1+((∣p∣)/(8x))−(1/(8x^3 ))))) =2x(√(1+((∣p∣)/(8x))−(1/(8x^3 ))))∼2x{1+(1/2)(((∣p∣)/(8x))−(1/(8x^3 )))} =2x{1+((∣p∣)/(16x))−(1/(16x^3 ))} =2x +((∣p∣)/8)−(1/(8x^2 )) ⇒ (...)−(....)∼2x−(1/2) +(1/(8x^2 ))−2x−((∣p∣)/8) +(1/(8x^2 )) ∼−(1/2)−((∣p∣)/8) ⇒ lim_(x→−∞) (....)=−(1/2)−((∣p∣)/8) =(1/4) ⇒−((∣p∣)/8) =(1/2)+(1/4) =(3/4) impossible with absolute value so there is a error in the Q.
wehave38x3−4x2+1=38x3(1−12x+18x3)=2x1−12x+18x3∼2x(1+12(−12x+18x3))(x→−∞)=2x{1−14x+116x3}=2x−12+18x2alsowehave38x3+∣p∣x2−1=38x3(1+∣p∣8x−18x3)=2x1+∣p∣8x−18x3∼2x{1+12(∣p∣8x−18x3)}=2x{1+∣p∣16x−116x3}=2x+∣p∣8−18x2⇒(…)−(….)∼2x−12+18x2−2x−∣p∣8+18x2∼−12−∣p∣8⇒limx→−∞(….)=−12−∣p∣8=14⇒−∣p∣8=12+14=34impossiblewithabsolutevaluesothereisaerrorintheQ.
Commented by mathmax by abdo last updated on 19/Feb/20
sorry error at line 4 we have ^3 (√(8x^3 +∣px∣^2 −1 ))=^3 (√(8x^3 +p^2 x^2 −1))=^3 (√(8x^3 (1+(p^2 /(8x))−(1/(8x^3 ))))) =2x(√(1+(p^2 /(8x))−(1/(8x^3 ))))∼2x{1+(1/2)((p^2 /(8x))−(1/(8x^3 )))} =2x+x((p^2 /(8x))−(1/(8x^3 ))) =2x+(p^2 /8)−(1/(8x^3 )) but the error still existing...
sorryerroratline4wehave38x3+∣px∣2−1=38x3+p2x2−1=38x3(1+p28x−18x3)=2x1+p28x−18x3∼2x{1+12(p28x−18x3)}=2x+x(p28x−18x3)=2x+p28−18x3buttheerrorstillexisting…
Answered by mr W last updated on 20/Feb/20
let p^2 =a>0 lim_(x→−∞) ((8x^3 −4x^2 +1))^(1/(3 )) −((8x^3 +∣px∣^2 −1))^(1/(3 )) =lim_(x→−∞) x(((8−(4/x)+(1/x^3 )))^(1/(3 )) −((8+(a/x)−(1/x^3 )))^(1/(3 )) ) =lim_(x→−∞) ((((8−(4/x)+(1/x^3 )))^(1/(3 )) −((8+(a/x)−(1/x^3 )))^(1/(3 )) )/(1/x)) =lim_(t→−0) ((((8−4t+t^3 ))^(1/(3 )) −((8+at−t^3 ))^(1/(3 )) )/t) =lim_(t→−0) (((2−(t/3)−(t^2 /(18))+o(t^2 ))−(2+((at)/(12))−((∣p∣^2 t^2 )/(288))+o(t^2 )) )/t) =−((1/3)+(a/(12)))≠(1/4) ! =−((1/3)+(a/(12)))≤−(1/3)
letp2=a>0limx→−∞8x3−4x2+13−8x3+∣px∣2−13=limx→−∞x(8−4x+1x33−8+ax−1x33)=limx→−∞8−4x+1x33−8+ax−1x331x=limt→−08−4t+t33−8+at−t33t=limt→−0(2−t3−t218+o(t2))−(2+at12−∣p∣2t2288+o(t2))t=−(13+a12)≠14!=−(13+a12)⩽−13
Commented by jagoll last updated on 19/Feb/20
if the limit value is equal to (1/3) then (1/3)+((∣p∣)/(12)) = (1/3) ⇒ p = 0 sir?
ifthelimitvalueisequalto13then13+∣p∣12=13⇒p=0sir?
Commented by mr W last updated on 19/Feb/20
you mean the limit =−(1/3)? then yes.
youmeanthelimit=−13?thenyes.

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