Hi everybody,
I must show that the following function accepts a limit in 1 :
$\displaystyle f(x)=\frac{x^{p+1}-(p+1)x+p}{(x-1)^2} (p\in \mathbb{N*})$
I know that i should factorize by (x-1) but how?
I need your help
Thank's anyway.
Hi
$\displaystyle x^{p+1}-(p+1)x+p = x^{p+1}-1-(p+1)x+p+1$
$\displaystyle x^{p+1}-(p+1)x+p = (x-1)\:\left(\sum_{k=0}^{p}x^k\right) -(p+1)(x-1)$
$\displaystyle x^{p+1}-(p+1)x+p = (x-1)\:\left(\sum_{k=0}^{p}x^k -(p+1)\right)$
$\displaystyle x^{p+1}-(p+1)x+p = (x-1)\:\left(\sum_{k=1}^{p}x^k -p\right)$
Let $\displaystyle g(x) = \sum_{k=1}^{p}x^k -p$
$\displaystyle g(1) = \sum_{k=1}^{p}1 -p = 0$
Therefore g(x) can be factored by (x-1)