Evaluate the limit:

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

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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`

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