Question Number 101615 by john santu last updated on 03/Jul/20

Commented bybramlex last updated on 03/Jul/20

i got −∞ sir

Commented byjohn santu last updated on 04/Jul/20

lim_(x→−∞)   ((16x^3 −1)/((√(16x^4 +16x^3 −1))−4x^2 ))  lim_(x→−∞)  ((16x^3 −1)/(−4x^2 (√(1+(1/x)−(1/(16x^2 ))))−4x^2 ))   lim_(x→−∞)  ((16x^3 −1)/(−4x^2 ((√(1+(1/x)−(1/(16x^2 ))))+1)))  lim_(x→−∞)  ((−x^3 (−16+(1/x^3 )))/(−4x^2 ((√(1+(1/x)−(1/(16x^2 ))))+1)))  lim_(x→−∞) ((x(−16+(1/x^3 )))/((√(1+(1/x)−(1/(16x^2 ))))+1)) = ∞ (JS⊛)  it correct ?

Commented byjohn santu last updated on 04/Jul/20

sorry i mistake write question.

Commented byjohn santu last updated on 04/Jul/20

it should be   lim_(x→−∞)  (√(16x^4 +16x^3 −1)) + 4x^2

Answered by bramlex last updated on 04/Jul/20

lim_(x→−∞)  (((16x^4 +16x^3 −1)−16x^4 )/(((16x^4 +16x^3 −1))^(1/ )  −4x^2 )) =  lim_(x→−∞) ((16x^3 −1)/((√((2x)^4 (1+(1/x)−(1/(16x^4 )))))−4x^2 )) =  lim_(x→−∞) ((−2x^2 (−8x+(1/(2x^2 ))))/(−2x^2 ((√(1+(1/x)−(1/(16x^4 ))))+2)))=  lim_(x→−∞) ((−8x+(1/(2x^2 )))/((√(1+(1/x)−(1/(16x^4 ))))+2)) = ∞

Commented bybramlex last updated on 03/Jul/20

oo sorry sir. you are right

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

lim_(x→−∞) (x(16+((16)/x)−(1/x^4 ))^(1/4) +4x^2 )=2x+4x^2 →∞

Answered by mathmax by abdo last updated on 05/Jul/20

let f(x) =(16x^4  +16x^3 −1)^(1/4)  +4x^2  ⇒f(x) =2x(1+(1/x)−(1/(16))x^4 )^(1/2)  +4x^2   ⇒f(x) ∼ 2x(1+(1/2)((1/x)−(x^4 /(16))))+4x^2   ⇒f(x) ∼ 2x{ 1+(1/(2x))−(x^4 /(32))} +4x^2  ⇒f(x) ∼2x+1 −(x^5 /(16)) +4x^2  ⇒  lim_(x→−∞) f(x) =lim_(x→−∞) −(x^5 /(16)) =+∞