# lim [ (x^3+1)^(1/3)  - x^(1/3) ] / [(x^2-1)^(1/2) +2 ] as x->- infinity  please? answer in textbook is (-1)

sciencesolve | Certified Educator

You need to substitute `-oo ` for x in the fraction to check if the limit is indeterminate such that:

`lim_(x-gt-oo) (root(3) (x^3+1)-root(3) x)/(sqrt(x^2-1)+2)=(-oo-oo)/oo = -oo/oo`

The limit is indeterminate, hence l'Hospital's theorem may help you such that:

`lim_(x-gt-oo) ((root(3) (x^3+1)-root(3) x)')/((sqrt(x^2-1)+2)')=lim_(x-gt-oo) (3x^2*(1/3)/(root(3)((x^3+1)^2)) - 1/(3root(3)(x^2)))/(x/sqrt(x^2+1))`

You need to force `x^3 ` factor to `(root(3)((x^3+1)^2))`  and `x^2`  factor to`sqrt(x^2+1)`  such that:

`lim_(x-gt-oo) (3x^2*(1/3)/(x^2root(3)((1+1/x^3)^2)) - 1/(3root(3)(x^2)))/(x/|x|sqrt(1+1/x^2))`

Reducing like terms yields:

`lim_(x-gt-oo) (1/(root(3)((1+1/x^3)^2)) - 1/(3root(3)(x^2)))/(1/sqrt(1+1/x^2))`

Substituting `-oo ` for x in limit yields:

`(1/(root(3)((1+1/(-oo))^2)) - 1/(3root(3)(oo)))/(1/sqrt(1+1/oo)) = (1/1 - 0)/(1/(-1))`  (notice that `1/oo -gt 0` )

`lim_(x-gt-oo) (root(3) (x^3+1)-root(3) x)/(sqrt(x^2-1)+2)= - 1`

Hence, evaluating the limit of function yields `lim_(x-gt-oo) (root(3) (x^3+1)-root(3) x)/(sqrt(x^2-1)+2)= - 1` .