Question Number 102036 by Study last updated on 06/Jul/20

lim_(x→0) ((((x^2 −1))^(1/5)  +((x+1))^(1/3) )/(((x−1))^(1/3)  +(√(x+1))))=?

Answered by john santu last updated on 06/Jul/20

L′Hopital   lim_(x→0)   ((((2x)/(5 (((x^2 −1)^4 ))^(1/5) ))+ (1/(3 (((x+1)^2 ))^(1/3) )))/((1/(3 (((x−1)^2 ))^(1/3) )) + (1/(2 (√((x+1))))))) =   ((1/3)/((1/3) +(1/2))) = (1/3) × (6/5) = (2/5)  (JS ⊛)

Answered by bemath last updated on 06/Jul/20

lim_(x→0) ((((x+1))^(1/3) −((1−x^2 ))^(1/5) )/((√(x+1))−((1−x))^(1/3) )) =  lim_(x→0) ((((x/3)+1)−(1−(x^2 /5)))/(((x/2)+1)−(1−(x/3)))) =  lim_(x→0) ((x((1/3)+(x/5)))/(x((1/2)+(1/3)))) = (1/3)×(6/5) = (2/5)

Commented by1549442205 last updated on 06/Jul/20

(1+x)^m =1+(m/(1!))x+((m(m−1))/(2!))x^2 +((m(m−1)(m−2))/(3!))x^3 +...

Commented byStudy last updated on 06/Jul/20

write the furmolla

Commented bybemath last updated on 06/Jul/20

maclaurin series

Commented byDwaipayan Shikari last updated on 06/Jul/20

Answered by Dwaipayan Shikari last updated on 06/Jul/20

lim_(x→0) ((((2x)/(5(x^2 −1)^(4/5) ))+(1/(3(x+1)^(2/3) )))/((1/(3(x−1)^(2/3) ))+(1/(2(x+1)^(1/2) ))))=((((2x)/5)+(1/3))/((1/3)+(1/2)))=(2/5)