Without what I said, Evaluate :
$\displaystyle \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$
Ok, do a change of variable. $\displaystyle u = x - 1$, so that as $\displaystyle x \to 1$, $\displaystyle u \to 0$. the limit becomes
$\displaystyle \lim_{u \to 0} \frac {(u + 1)^5 - 1}u$
now expand the top, cancel the $\displaystyle u$'s, evaluate. do you like that one?
You don't find this by long division you observe that $\displaystyle (x-1)(x^{n-1}+x^{n-2}+... +1)$ gives a telescoping sum. Or equivalently:
$\displaystyle x^n-1=[x^n+x^{n-1}+ ... +x] - [x^{n-1}+x^{n-2}+ ... +1] $
......... $\displaystyle = (x^n-x^{n-1}) + (x^{n-1}-x^{n-2})+ ... + (x-1)$
......... $\displaystyle = x^{n-1}(x-1) + x^{n-2}(x-1) + ... + (x-1)$
......... $\displaystyle = (x-1)[x^{n-1} + x^{n-2} + ... +1]$
CB
I think I can do it without what you said but using the ideas from calculus. Your limit represent the slope of the tangent of $\displaystyle y = x^5$ at $\displaystyle x = 1$. Let
$\displaystyle \lim_{x \to 1} \frac{x^5-1}{x-1} = m$.
The tangent line is $\displaystyle y-1 = m(x-1)$ or $\displaystyle y = mx -m +1$
The intersection of the tangent line and the quintic
$\displaystyle x^5 = mx - m + 1$ or $\displaystyle x^5 -mx + m -1 = 0 .\;\;(1.1)$
There should be two intersection points, at $\displaystyle x = 1$ and $\displaystyle x = a$ (some $\displaystyle a$) where the first will be a double root so eqn. (1.1) has the form
$\displaystyle
(x-1)^2(x-a)(x^2+px+q) = 0 \;\;(1.2)
$ for some $\displaystyle p$ and $\displaystyle q$.
Expanding and equating the coefficients of $\displaystyle (1.1)$ and $\displaystyle (1.2)$ gives
$\displaystyle (2.1)\;\;m + aq - 1 = 0, $
$\displaystyle (2.2)\;\;-m + ap - 2aq - q = 0,$
$\displaystyle
(2.3)\;\;aq - 2ap + 2q + a - p = 0,
$
$\displaystyle
(2.4)\;\;ap + 2p - q - 2a - 1 =0,
$
$\displaystyle (2.5)\;\;a-p+2 = 0.$
If we add/substract in the following order
$\displaystyle
(2.1) - (2.2)- 3 \times (2.3) - 5 \times (2.4) - 7 \times (2.5) = 0
$
gives
$\displaystyle 2m - 10 = 0$ or $\displaystyle m = 5$, our desired limit.