1. ## Limit problem

I´m having trouble with this limit problem, would appreciate help.

$lim_{x\to\1}\frac{x+x^2+...+x^n-n}{x-1}$

2. ## Re: Limit problem

Originally Posted by HK47
I´m having trouble with this limit problem, would appreciate help.
$lim_{x\to\1}\frac{x+x^2+...+x^n-n}{x-1}$
Since both numerator and denominator approach zero you can apply L'Hopital's Rule.

If you are not allowed to use that then the algebra becomes very messy.
Note for each positive integer $k$, $(x-1)$ is a factor of $x^k-1$.

3. ## Re: Limit problem

Yeah, cant use the L´Hopital quite yet. My math textbook sure likes to stump me with these exercises, which can not be solved by provided means. At least I can post them here, so I dont have to rack my brains indefinitely.

4. ## Re: Limit problem

Originally Posted by HK47
Yeah, cant use the L´Hopital quite yet. My math textbook sure likes to stump me with these exercises, which can not be solved by provided means. At least I can post them here, so I dont have to rack my brains indefinitely.
Take note:
\begin{align*} \frac{{x + {x^2} + \cdots + {x^n} - n}}{{x - 1}} &= \frac{{(x - 1) + ({x^2} - 1) + \cdots + ({x^n} - 1)}}{{x - 1}} \\&= \frac{{\sum\limits_{k = 1}^n {({x^k} - 1)} }}{{x - 1}}\\ &= \sum\limits_{k = 1}^n {\left( {\sum\limits_{j = 0}^{k - 1} {{x^j}} } \right)} \end{align*}

5. ## Re: Limit problem

So the limit is infinity. Thank you! Didn´t think to factor the n of all things.

6. ## Re: Limit problem

Originally Posted by HK47
So the limit is infinity.

No! Actually the limit is $\frac{n(n+1)}{2}$.

7. ## Re: Limit problem

Oh okay, right, n isnt infinity

8. ## Re: Limit problem

Originally Posted by HK47
Oh okay, right, n isnt infinity
n is never infinity, because infinity is not a number.