Question Number 101610 by bramlex last updated on 03/Jul/20

Answered by john santu last updated on 04/Jul/20

L′Hopital  ⇔ lim_(x→0)  ((x^2 (√(x^3 +5)))/(sin 2x)) = lim_(x→0)  (x/(sin 2x)) ×lim_(x→0)  x(√(x^3 +5))  = 0 (JS⊛)

Answered by Dwaipayan Shikari last updated on 04/Jul/20

lim_(x→0) (((1/3)∫_0 ^x 2u^2 du)/(sin^2 x))=lim_(x→0) (((2/9)[u^3 ]_(√5) ^(√(x^3 +5)) )/(sin^2 x))=lim_(x→0) (((2/9)[(x^3 +5)^(2/3) −5^(2/3) ])/(x^2 sin^2 x)).x^2     lim_(x→0) (((2/9)[(x^3 +5)^(2/3) −5^(2/3) ])/((x^3 +5)−5)).((x^2 x)/x^2 )=(2/9).(2/3) 5^(1/3) .x=   0  (DS)  ■★{assuming sinx=x}