Question Number 104655 by bobhans last updated on 23/Jul/20

lim_(x→a) (((√(2x+a))−(√(3x)))/((√(x+3a))−2(√x))) =?    [a≠0]

Answered by bramlex last updated on 23/Jul/20

lim_(x→a)   (([(1/(√(2x+a)))−(3/(2(√(3x))))])/([(1/(2(√(x+3a))))−(1/(√x))])) = (((1/(√(3a)))−(3/(2(√(3a)))))/((1/(4(√a)))−(1/(2(√a)))))  = {(2/(2(√(3a))))−(3/(2(√(3a)))) }× (−4(√a) )  = −(1/(2(√(3a)))) × (−4(√a) )  = (2/(√3)) ★   [ for a > 0 ]

Commented bymalwaan last updated on 23/Jul/20

(2/(3(√3)))

Answered by john santu last updated on 23/Jul/20

set x = a+p  lim_(p→0)  (((√(2p+3a)) −(√(3p+3a)))/((√(p+4a))−(√(4p+4a)))) =  lim_(p→0) (((√(3a)) {(√(1+((2p)/(3a))))−(√(1+(p/a) )) })/((√(4a)) {(√(1+(p/(4a))))−(√(1+(p/a)))}))  = ((√3)/2) × lim_(p→0) (((1+(p/(3a)))−(1+(p/(2a))))/((1+(p/(8a)))−(1+(p/(2a)))))  = ((√3)/2)× lim_(p→0)  (((−(p/(6a))))/((−((3p)/(8a)))))  = ((√3)/2) × ((1/(6a)))×(((8a)/3))= ((8(√3))/(2×3×6))  = ((2(√3))/9)

Answered by bemath last updated on 23/Jul/20

standart method  lim_(x→a)  (((√(x+3a))+2(√x))/((√(2x+a))+(√(3x)))) × lim_(x→a)  ((2x+a−3x)/(x+3a−4x)) =  ((4(√a))/(2(√(3a)))) × lim_(x→a)  ((a−x)/(3(a−x))) =  (2/(√3)) × (1/3) = (2/(3(√3))) = ((2(√3))/9) ■

Answered by Dwaipayan Shikari last updated on 23/Jul/20

lim_(x→a) (((√(3x))−(√(2x+a)))/(2(√x)−(√(x+3a))))=lim_(x→a) (((√(3x))−(√(2x+a)))/(3x−2x−a)).((4x−x−3a)/(2(√x)−(√(x+3a)))).((3x−2x−a)/(4x−x−3a))  (1/2)(2x+a)^(−(1/2)) .2(√(x+3a)).((x−a)/(3(x−a)))  (1/(√(3a))).2(√a).(1/3)=(2/(√3)).(1/3)=(2/(3(√3)))★■