Question Number 104533 by 175mohamed last updated on 22/Jul/20

find :  lim_(x→1) (((x+2)^2 +(x+1)^3 −17)/((√(x+3))−((x+7))^(1/3) ))

Answered by Dwaipayan Shikari last updated on 22/Jul/20

lim_(x→1) ((2(x+2)+3(x+1)^2 )/((1/2).(1/(√(x+3)))−(1/3)(x+7)^(−(2/3)) ))=((18)/((1/4)−(1/(12))))=108

Answered by OlafThorendsen last updated on 22/Jul/20

x = u+1  lim_(u→0) (((u+3)^2 +(u+2)^3 −17)/((√(u+4))−((u+8))^(1/3) ))  lim_(u→0) ((9(1+(u/3))^2 +8(1+(u/2))^3 −17)/(2(√(1+(u/4)))−2((1+(u/8)))^(1/3) ))  lim_(u→0) ((9(1+((2u)/3))+8(1+((3u)/2))−17)/(2(1+(u/8))−2(1+(u/(24)))))  lim_(u→0) ((9(((2u)/3))+8(((3u)/2)))/(2((u/8))−2((u/(24)))))  lim_(u→0) ((6u+12u)/((u/4)−(u/(12))))  lim_(u→0) ((18u)/(u/6)) = 18×6 = 108

Commented bybobhans last updated on 22/Jul/20

(u/4)−(u/(12)) = ((3u)/(12))−(u/(12)) = ((2u)/(12)) = (u/6)

Answered by bramlex last updated on 22/Jul/20

L′Hopital rule   lim_(x→1) ((2(x+2)+3(x+1)^2 )/((1/(2(√(x+3))))−(1/(3 (((x+7)^2 ))^(1/3) )))) =  ((6+12)/((1/4)−(1/(3.4)))) = ((18)/(((1/6)))) = 108 □