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