We will use the following formula

`x^n-1=(x-1)(x^(n-1)+x^(n-2)+x^(n-3)+cdots+x+1)`

Now our limit

`lim_(x->1)(x^n-1)/(x-1)`

becomes

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

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

`1+1+1+cdots+1=n`

**Hence the solution is ** `lim_(x->1)(x^n-1)/(x-1)=n`.

Alternatively you could use L'Hospital's rule.

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

`lim_(x->1)nx^(n-1)=n`

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

If we directly substitute x = 1 in the expression `(x^n-1)/(x-1)` , the result is `(1-1)/(1-1) = 0/0` . The value of `0/0` is not defined or indeterminate. In limits where the expression takes on the form `0/0` or `oo/oo` or `0^0` among many others L'Hospital's rule can be used to find the limit.

This involves replacing the numerator and the denominator with their derivatives.

This gives:

`lim_(x->1) (n*x^(n-1) - 0)/(1-0)`

Now substituting x = 1 gives n*1 = n

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