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Question Number 130843 by EDWIN88 last updated on 29/Jan/21

Without L′Ho^� pital lim_(x→1) ((((x^3 +7))^(1/3) − (√(x^2 +3)))/(x−1)) ?

WithoutL′Hopital^limx→1x3+73−x2+3x−1?

Answered by mathmax by abdo last updated on 29/Jan/21

let f(x)=(((x^3 +7)^(1/3) −(x^2 +3)^(1/2) )/(x−1)) we do the changement x−1=t ⇒ f(x)=f(t+1) =((((t+1)^3 +7)^(1/3) −((t+1)^2 +3)^(1/2) )/t) =(((t^3 +3t^2 +3t +8)^(1/3) −(t^2 +2t+4)^(1/2) )/t) =((8^(1/3) (1+(1/8)(t^3 +3t^2 +3t))^(1/3) −4^(1/2) (1+(1/4)(t^2 +2t))^(1/2) )/t)( t→0) ⇒ f(t+1)∼((2{1+(1/(24))(t^3 +3t^2 +3t)}−2(1+(1/8)(t^2 +2t)))/t) ⇒f(t+1)∼(((1/(12))(t^3 +3t^2 +3t)−(1/4)(t^2 +2t))/t) ⇒f(t+1)∼(1/(12))(t^2 +3t+3)−(1/4)(t+2) ⇒ lim_(t→0) f(t+1) =(3/(12))−(2/4)=(1/4)−(1/2)=−(1/4) ⇒ lim_(x→1) f(x)=−(1/4)

letf(x)=(x3+7)13−(x2+3)12x−1wedothechangementx−1=t⇒f(x)=f(t+1)=((t+1)3+7)13−((t+1)2+3)12t=(t3+3t2+3t+8)13−(t2+2t+4)12t=813(1+18(t3+3t2+3t))13−412(1+14(t2+2t))12t(t→0)⇒f(t+1)∼2{1+124(t3+3t2+3t)}−2(1+18(t2+2t))t⇒f(t+1)∼112(t3+3t2+3t)−14(t2+2t)t⇒f(t+1)∼112(t2+3t+3)−14(t+2)⇒limt→0f(t+1)=312−24=14−12=−14⇒limx→1f(x)=−14

Commented by EDWIN88 last updated on 30/Jan/21

yes..thank you

yes..thankyou

Commented by mathmax by abdo last updated on 30/Jan/21

you are welcome

youarewelcome

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