Question Number 148439 by liberty last updated on 28/Jul/21

lim_(x→2) (((√(x+2)) ((x+6))^(1/3) −x^2 )/(x−2)) =?

Answered by dumitrel last updated on 28/Jul/21

lim_(x→2) (((√(x+2))(((x+6))^(1/3) −2))/(x−2))+lim_(x→2) ((2((√(x+2))−2))/(x−2))−lim_(x→2) ((x^2 −4)/(x−2))= lim_(x→2) (((√(x+2))(x−2))/((x−2)((((x+6)^2 ))^(1/3) +2((x+2))^(1/3) +4))+lim_(x→2) ((2(x−2))/((x−2)((√(x+2))+2)))+lim_(x→2) (((x−2)(x+2))/(x−2)))= =(2/(4+4+4))+(2/(2+2))−4=−((10)/3)

Answered by EDWIN88 last updated on 28/Jul/21

Answered by mathmax by abdo last updated on 28/Jul/21

f(x)=(((√(x+2))(x+6)^(1/3) −x^2 )/(x−2)) changement x−2=t give f(x)=g(t)=(((√(t+4))(t+8)^(1/3) −(t+2)^2 )/t) =(((√(t+4))(t+8)^(1/3) −t^2 −4t−4)/t)=((2(√(1+(t/4)))(^3 (√8)(1+(t/8))^(1/3) −t^2 −4t−4)/t) ⇒g(t)∼((4(1+(t/8))(1+(t/(24)))−t^2 −4t−4)/t) =((4(1+(t/(24))+(t/8)+(t^2 /(8.24)))−t^2 −4t−4)/t) =(((t/6)+(t/2)+(t^2 /(48))−t^2 −4t)/t)=(1/6)+(1/2)−4+((t/(48))−t) ⇒ lim_(t→0) g(t)=(2/3)−4 =−((10)/3) ⇒lim_(x⌣2) f(x)=−((10)/3)