# Math Help - Without L.R.,Long division,definition of derivative and eplison-delta prove, evaluate

1. ## Without L.R.,Long division,definition of derivative and eplison-delta proof, evaluate

Without what I said, Evaluate :
$\lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$

2. Originally Posted by General
Without what I said, Evaluate :
$\lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$
factor: $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$, the bad guys cancel, evaluate at 1 and you're done.

3. Originally Posted by Jhevon
factor: $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$, the bad guys cancel, evaluate at 1 and you're done.
How did you factor it ?
1 is zero of $x^5-1$
then (x-1) is a factor .. and you find $x^4+x^3+x^2+x+1$ by long division
which is not allowed to use.

4. Originally Posted by General
How do you factor it ?
It's a known fact that $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1})$ when $n$ is an odd natural number

5. Originally Posted by Jhevon
It's a known fact that $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1})$ when $n$ is an odd natural number
OK.
Waiting for the other solutions.
I want to see another ideas

Thanks.

6. Originally Posted by General
OK.
Waiting for the other solutions.
I want to see another ideas

Thanks.
Ok, do a change of variable. $u = x - 1$, so that as $x \to 1$, $u \to 0$. the limit becomes

$\lim_{u \to 0} \frac {(u + 1)^5 - 1}u$

now expand the top, cancel the $u$'s, evaluate. do you like that one?

7. Originally Posted by Jhevon
Ok, do a change of variable. $u = x - 1$, so that as $x \to 1$, $u \to 0$. the limit becomes

$\lim_{u \to 0} \frac {(u + 1)^5 - 1}u$

now expand the top, cancel the $u$'s, evaluate
Its the same idea
Thank you.

8. Originally Posted by Jhevon
Ok, do a change of variable. $u = x - 1$, so that as $x \to 1$, $u \to 0$. the limit becomes

$\lim_{u \to 0} \frac {(u + 1)^5 - 1}u$

now expand the top, cancel the $u$'s, evaluate. do you like that one?
I wouldn't recommend that if having $\frac{x^{n}-1}{x-1},$ you wouldn't expand.

9. Originally Posted by Krizalid
I wouldn't recommend that if having $\frac{x^{n}-1}{x-1},$ you wouldn't expand.
Haha, point taken. for arbitrary n i would use one of the methods we have been restricted from using. here it's ok though

do you have another method for arbitrary (large) n values?

10. Originally Posted by General
How did you factor it ?
1 is zero of $x^5-1$
then (x-1) is a factor .. and you find $x^4+x^3+x^2+x+1$ by long division
which is not allowed to use.
You don't find this by long division you observe that $(x-1)(x^{n-1}+x^{n-2}+... +1)$ gives a telescoping sum. Or equivalently:

$x^n-1=[x^n+x^{n-1}+ ... +x] - [x^{n-1}+x^{n-2}+ ... +1]$

......... $= (x^n-x^{n-1}) + (x^{n-1}-x^{n-2})+ ... + (x-1)$

......... $= x^{n-1}(x-1) + x^{n-2}(x-1) + ... + (x-1)$

......... $= (x-1)[x^{n-1} + x^{n-2} + ... +1]$

CB

11. 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 $y = x^5$ at $x = 1$. Let

$\lim_{x \to 1} \frac{x^5-1}{x-1} = m$.

The tangent line is $y-1 = m(x-1)$ or $y = mx -m +1$

The intersection of the tangent line and the quintic

$x^5 = mx - m + 1$ or $x^5 -mx + m -1 = 0 .\;\;(1.1)$

There should be two intersection points, at $x = 1$ and $x = a$ (some $a$) where the first will be a double root so eqn. (1.1) has the form

$
(x-1)^2(x-a)(x^2+px+q) = 0 \;\;(1.2)
$
for some $p$ and $q$.

Expanding and equating the coefficients of $(1.1)$ and $(1.2)$ gives

$(2.1)\;\;m + aq - 1 = 0,$
$(2.2)\;\;-m + ap - 2aq - q = 0,$
$
(2.3)\;\;aq - 2ap + 2q + a - p = 0,
$

$
(2.4)\;\;ap + 2p - q - 2a - 1 =0,
$

$(2.5)\;\;a-p+2 = 0.$

If we add/substract in the following order

$
(2.1) - (2.2)- 3 \times (2.3) - 5 \times (2.4) - 7 \times (2.5) = 0
$

gives

$2m - 10 = 0$ or $m = 5$, our desired limit.

12. Originally Posted by Jhevon
It's a known fact that $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \cdots + ab^{n - 2} + b^{n - 1})$ when $n$ is an odd natural number
$a^n - b^n = (a-b)\left(\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\right), \forall \ n\in \mathbb{N}$.

It's $a^n + b^n = (a+b)\left(\sum_{k=0}^{n-1}(a^{n-k-1})(-b)^k\right)$ that only holds when n is an odd positive integer.