You should substitute 1 for x in the equation under limit such that:

`lim_(x->1) (x^3-1)/|x^3-1| = (1-1)/|1-1| = 0/0`

You need to evaluate the side limits such that:

`lim_(x->1, x<1) (x^3-1)/|x^3-1| = lim_(x->1, x<1) (x^3-1)/(1 - x^3) = 0/0`

Since evaluating the limit yields an indetermination, you may use l'Hospital's theorem such that:

`lim_(x->1,x<1) (x^3-1)/(1 - x^3) = lim_(x->1) ((x^3-1)')/((1 - x^3)')`

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

Reducing like terms yields

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

`lim_(x->1, x>1) (x^3-1)/|x^3-1| = lim_(x->1, x>1) (x^3-1)/(x^3-1) = 0/0`

`lim_(x->1, x>1) (x^3-1)/(x^3-1) = lim_(x->1,x>1) ((x^3 - 1)')/((x^3 - 1)')`

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

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

Conclusion: `lim_(x->1, x>1) = 1 != -1 lim_(x->1, x<1)`

**Hence, evaluating side limits yields that they do not have equal values, thus, that there is no ordinary limit for the given function, under the given condition.**