Question Number 103366 by Study last updated on 14/Jul/20

Answered by mathmax by abdo last updated on 14/Jul/20

let f(x) =((x^(1/4) −5^(1/{) )/((lnx)^(1/8) −(ln5)^(1/8) ))  let find lim_(x→5) f(x) we do the changement  x =5−t ⇒f(x) =g(t) =(((5−t)^(1/4) −5^(1/4) )/((ln(5−t))^(1/8) −(ln5)^(1/8) ))   (t→0)  =((5^(1/4) {(1−(t/5))^(1/4) −1})/((ln(5)+ln(1−(t/5)))^(1/8) −(ln5)^(1/8) ))  ⇒g(t) ∼((5^(1/4) {1−(t/(20))−1})/((ln(5)−(t/5))^(1/8) −(ln5)^(1/8) )) =((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {(1−(t/(5ln5)))^(1/8) −1}))  ∼((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {1−(t/(40ln5))−1})) →(5^(1/4) /((ln5)^(1/8) ))×((40ln5)/(20)) =((2 ln5 ×5^(1/4) )/((ln5)^(1/8) ))  =2.5^(1/4)  (ln5)^(7/8)   ⇒lim_(x→5) f(x)= 2(^4 (√5))(ln5)^(7/8)

Answered by bemath last updated on 14/Jul/20

(1) lim_(x→5)   (((x)^(1/4) −(5)^(1/4) )/(((lnx))^(1/8)  −((ln 5))^(1/8) )) =   lim_(x→5)  ((1/(4 x^(3/4) ))/(1/(8x (ln x)^(7/8) ))) = lim_(x→5)  ((8x (ln x)^(7/8) )/(4x^(3/4) ))  = ((10.(ln 5)^(7/8) )/5^(3/4) ) ■

Answered by bemath last updated on 14/Jul/20

(2) lim_(x→3)  (((e^x )^(1/6) −(e^3 )^(1/6) )/((x)^(1/3) −(3)^(1/3) )) =   lim_(x→3)  ((((e^x )^(−5/6) .e^x )/6)/(1/(3 (x^2 )^(1/3)  ))) =   lim_(x→3)  ((3 (x^2 )^(1/3) )/6) ×e^(x/6)  = (((9)^(1/3)  .e^(1/2) )/2)   = (((√e) .(9)^(1/3) )/2) ■

Answered by mathmax by abdo last updated on 15/Jul/20

let g(x) =((e^(x/6) −e^(1/2) )/(x^(1/3) −3^(1/3) ))  we do the cha7gement t =x−3 (so t→0) ⇒  g(x) =((e^((t+3)/6) −e^(1/2) )/((t+3)^(1/3) −3^(1/3) )) =((e^(1/2) {e^(t/6) −1})/(3^(1/3) { (1+(t/3))^(1/3) −1})) =ϕ(t) ⇒  ϕ(t) ∼(e^(1/2) /3^(1/3) ){ ((1+(t/6)−1)/(1+(t/9)−1))} →(e^(1/2) /3^(1/3) )×(9/6) =(3/2)×((√e)/((^3 (√3))))(t→0) ⇒  lim _(x→3) g(x) =((3(√e))/(2(^3 (√3))))