Without what I said, Evaluate :

$\displaystyle \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$

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- Jan 13th 2010, 12:19 PMGeneralWithout L.R.,Long division,definition of derivative and eplison-delta proof, evaluate
Without what I said, Evaluate :

$\displaystyle \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$ - Jan 13th 2010, 12:26 PMJhevon
- Jan 13th 2010, 12:27 PMGeneral
- Jan 13th 2010, 12:32 PMJhevon
- Jan 13th 2010, 12:33 PMGeneral
- Jan 13th 2010, 12:41 PMJhevon
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? - Jan 13th 2010, 12:42 PMGeneral
- Jan 13th 2010, 01:21 PMKrizalid
- Jan 13th 2010, 01:33 PMJhevon
- Jan 13th 2010, 02:11 PMCaptainBlack
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 - Jan 14th 2010, 07:21 AMJester
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. - Jan 14th 2010, 06:24 PMI4talent