Answer 1

`lim_(y -> 1) (2y - 2)/(root(3)(7 + y) - 2)`

= `lim_(y -> 1) (2(y - 1))/((7 + y)^(1/3) - 8^(1/3))    ...[because 2 = (2^3)^(1/3) = 8^(1/3)]`

= `lim_(y -> 1) 2/(((y + 7)^(1/3) - 8^(1/3))/(y - 1)`

= `(lim_(y -> 1)  2)/(lim_(y -> 1) ((y + 7)^(1/3) - 8^(1/3))/((y + 7) - 8)`

Let y + 7 = x
As y → 1, x → 8

= `2/(lim_(x -> 8)  (x^(1/3) - 8^(1/3))/(x - 8))`

= `2/(1/3(8)^((-2)/2))      ...[lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `2(3)*(8)^(2/3)`

= `6(2^3)^(2/3)`

= 6 x (2)²
= 24