Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$? In the back of De Souza (Berkeley Problems in Mathematics, page 305), it says:

For $x \neq 1$,
  $$
f(x) = (x^n-1)/(x-1) = x^{n-1} + \cdots + 1
$$
  so $f(1) = n$.

The expansion for $x \neq 1$ obviously follows from the definition of a partial geometric series. But since it requires $x \neq 1$, how can it follow that $f(1)=n$?
Edit: $f$ is defined as above as a polynomial. I think this changes the question since if $f$ is given as a polynomial, it must then be continuous. The only way for $f$ to be continuous is to define $f(1)=n$. Someone else can confirm if this is correct reasoning.

so must be replaced by and, then we can say "[...] so that $f$ is continuous everywhere.".
