What is limit of `(x^3 - 1)/(x-1)`  if x goes to 1?

2 Answers | Add Yours

ishpiro's profile pic

ishpiro | College Teacher | (Level 1) Educator

Posted on

To calculate `lim_(x->1) (x^3-1)/(x-1)` , factor the numerator of the fraction as a difference of two cubes:

`(x^3 - 1)/(x - 1) = ((x-1)(x^2 + x + 1))/(x-1) = x^2 + x + 1`

The resultant function is continuous at x = 1, so the limit can be evaluated by plugging in x = 1:

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

This limit equals 3.

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The limit `lim_(x->1) (x^3 - 1)/(x - 1)` has to be determined.

If we substitute x = 1, the numerator x^3 - 1 = 1 - 1 = 0 and the denominator x - 1 =  1 - 1  = 0. The result is the indeterminate form `0/0` . This allows us to use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

The limit is now:

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

Substituting x = 1 gives the result 3.

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

We’ve answered 315,468 questions. We can answer yours, too.

Ask a question