## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to1}\frac{x^n-1}{x-1}=n.$$
To solve this limit we will use the factor formula $x^n-1=(x-1)(x^{n-1}+x^{n-2}+\ldots x+1)$: $$\lim_{x\to1}\frac{x^n-1}{x-1}=\lim_{x\to1}\frac{(x-1)(x^{n-1}+x^{n-2}+\ldots x+1)}{x-1}=\lim_{x\to1}(x^{n-1}+x^{n-2}+\ldots x+1)=1^{n-1}+1^{n-2}+\ldots+1+1=n\cdot 1=n.$$ Note that in the last step we used the fact that we had $n$ terms all equal to one ($x^{n-1}$ to $x$ is $n-1$ terms and plus the extra $1$ at the end is $n$ terms).