Question

Question #90e9c

Answer 1

6

Explanation:

The answer is found using L'hopital's rule, which states that given continuous and differentiable `f(x) & g(x), lim_(x->x_0) f(x)/g(x) = lim_(x->x_0)(f'(x))/(g'(x))`.

Here, we have `f(x) = x^3-1 -> f'(x) = 3x^2, g(x) = sqrt(2x+2)-2 = (2x+2)^(1/2)-2 -> g'(x) = 1/2 *2 (2x+2)^-(1/2) = 1/sqrt(2x+2)`.

Yielding...

`lim_(x->x_0) f(x)/g(x) = lim_(x->x_0)(f'(x))/(g'(x)) -> lim_(x->1) (x^3-1)/(sqrt(2x+2)-2) = lim_(x->1) (3x^2)/(1/(2x+2)) = lim_(x->1) (3x^2)(sqrt(2x+2))`

From here, we simply plug in x=1...

`= 3(1^2)sqrt(2+2) = 3*sqrt(4) = 3*2 = 6`