*Originally posted by SwissGambit*

**Here's something I couldn't look up:
**

The limit, as x goes to 1, of...

(x - x^(n-1))/(1-x)

...is n-2, but why?

OK, here is as spiffy a solution as I could find.

Convert the equation

f(x) = (x-x^(n-1))/(1-x)

= x(x^(n-2)-1))/(x-1)

Let's take the denominator off for a bit, so it's easier to read.

x(x^(n-2)-1))

Now we start adding and subtracting the same term from the numerator over and over to factor out x-1 from every term. This will create new terms, but we'll do the same trick on them until EVERY term has an (x-1) in it.

x(x^(n-2) - x^(n-3) + x^(n-3) - 1)

x(x^(n-3)(x-1) + x^(n-3) - x^(n-4) + x^(n-4) - 1)

x(x^(n-3)(x-1) + x^(n-4)(x-1) + x^(n-4) - 1)

Keep doing this until you have dropped x's exponent down to one and you're left with an x-1 at the end.

x(x^(n-3)(x-1) + x^(n-4)(x-1) + ... + x(x-1) + x-1)

Now add the denominator back in:

x((x^(n-3)(x-1) + x^(n-4)(x-1) + ... + x(x-1) + x-1))/(x-1)

Cancel out all the (x-1)'s:

x(x^(n-3) + x^(n-4) + ... + x + 1)

Multiply the outer x through:

f(x) = x^(n-2) + x^(n-3) + ... + x^2 + x

Now that the denominator is gone, there is no more divide by 0 problem. With x=1 in each term, and n-2 total terms, the limit of f(x) as x approaches 1 is

**n-2**.