Question Number 115027 by bemath last updated on 23/Sep/20

If lim_(x→3)  ((17 ((ax+3))^(1/(3 ))  +b)/(x−3)) = ((136)/(27))  then 8a+b = ?

Answered by bobhans last updated on 23/Sep/20

limit form (0/0). numerator must be = 0  (1) 17 ((3a+3))^(1/(3 ))  +b =0;  b ⇒−17 ((3a+3))^(1/(3 ))   (2) lim_(x→3)  ((17 ((ax+3))^(1/3)  −17 ((3a+3))^(1/(3 )) )/(x−3)) =((136)/(27))         lim_(x→3)  ((((ax+3))^(1/(3 ))  −((3a+3))^(1/(3 )) )/(x−3)) = (8/(27))  set x=3+r ; r→0      lim_(r→0)  ((((ar+3a+3))^(1/(3 ))  −((3a+3))^(1/(3 )) )/r) = (8/(27))  remember (p)^(1/3)  −(q)^(1/(3 ))  = ((p−q)/( (p^2 )^(1/(3 )) +((pq))^(1/(3 ))  +(q^2 )^(1/(3 )) ))   lim_(r→0)  (((ar+3a+3)−(3a+3))/(r ((((ar+3a+3)^2 ))^(1/(3 )) +(((3a+3)(ar+3a+3)))^(1/(3 )) +(((3a+3)^2 ))^(1/(3 )) )) =(8/(27))  lim_(r→0)  ((ar)/(r((((ar+3a+3)^2 ))^(1/(3 )) +(((ar+3a+3)(3a+3)))^(1/(3 )) +(((3a+3)^2 ))^(1/(3 )) )))=(8/(27))  ⇔ (a/(3 (((3a+3)^2 ))^(1/(3 )) )) = (8/(27)) ;  by inspection we get a = 8 , check (8/(3 (((24+3))^(1/(3 )) )^2 ))  = (8/(3×9)) = (8/(27)). Then b = −17 ((24+3))^(1/3)  =−51  ∴ 8a+b = 64−51=13