`lim_(xrarroo) root3(x)(root3((x+1)^(2))-root3((x-1)^(2)))=`

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. 1
D. `(4)/(3)`

Answer

D

Explanation

`underset(xrarroo)(lim)root3x(root3((x+1)^(2))-root3((x-1)^(2)))`
`=underset(xrarroo)(lim)x^(1//3){(x+1)^(1//3)+(x-1)^(1//3)}{(x+1)^(1//3)-(x-1)^(1//3)}`
`" "underset(xrarroo)(lim)(x^(1//3){(x+1)^(1//3)+(x-1)^(1//2)}2)/({(x+1)^(2//3)+(x^(2)-1)^(1//3)+(x-1)^(2//3)})`
`" "=underset(xrarroo)(lim)(2.x^(2//3){(1+(1)/(x))^(1//3)+(1-(1)/(x))^(1//3)})/(x^(2//3){(1+(1)/(x))^(2//3)+(1-(1)/(x^(2)))^(1//3)+(1-(1)/(x))^(2//3)})=(4)/(3)`