How to evaluate the following limit? $\lim\limits_{x\to1}\frac{x^n+x^{n-1}+x^{n-2}+\cdots+x-n}{x-1}.$

Question

How to evaluate the following limit?

$$\lim_{x\to1}\dfrac{x^n+x^{n-1}+x^{n-2}+\cdots+x-n}{x-1}.$$

I could simplify but couldn't proceed further

Answer 1

HINT:

Set $x-1=y\iff x=y+1, y\to0$ as $x\to1$

and $x^m=(1+y)^m=1+my+O(m^2)$

$$\implies\sum_{m=1}^n x^m=\sum_{m=1}^n[1+my+O(y^2)]=\sum_{m=1}^n1+y\sum_{m=1}^nm+O(y^2)$$

Answer 2

You can tackle this problem using synthetic division and a simple series. First let's do the following division: $$\frac{x^n + x^{n-1} + x^{n-2} + \cdots + x - n}{x-1}$$

That would look like

 (x-1)    | x^n     x^n-1    x^n-2    ...    x    1
----------|  1        1        1      ...    1    -n
   1      |  0        1        2      ...    n-1  n       
          |  1        2        3      ...    n    0

We can therefor see that

$$\frac{x^n + x^{n-1} + x^{n-2} + \cdots + x - n}{x-1} = x^{n-1} + 2x^{n-2} + 3x^{n-3} + \cdots + n$$

Plugging in $x = 1$ we get:

$$\lim_{x\to1}\frac{x^n + x^{n-1} + x^{n-2} + \cdots + x - n}{x-1} = 1 + 2 + 3 + \cdots + n = \sum_{i=1}^n i = \frac{n(n+1)}{2}$$

Edit: Oops! I had a $n$ in the denominator instead of a 2. Thanks, Timbuc.