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

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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.**

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`