text: # Without L.R.,Long division,definition of derivative and eplison-delta prove, evaluate
  
  • Jan 13th 2010, 01:19 PM
  General
  Without L.R.,Long division,definition of derivative and eplison-delta proof, evaluate
  Without what I said, Evaluate :
  $\lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$
  • Jan 13th 2010, 01:26 PM
  Jhevon
  Quote:
  
  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.
  • Jan 13th 2010, 01:27 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 01:32 PM
  Jhevon
  Quote:
  
  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
  • Jan 13th 2010, 01:33 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 01:41 PM
  Jhevon
  Quote:
  
  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?
  • Jan 13th 2010, 01:42 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 02:21 PM
  Krizalid
  Quote:
  
  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. :D
  • Jan 13th 2010, 02:33 PM
  Jhevon
  Quote:
  
  Originally Posted by Krizalid
  I wouldn't recommend that if having $\frac{x^{n}-1}{x-1},$ you wouldn't expand. :D
  
  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?
  • Jan 13th 2010, 03:11 PM
  CaptainBlack
  Quote:
  
  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
  • Jan 14th 2010, 08:21 AM
  Jester
  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.
  • Jan 14th 2010, 07:24 PM
  I4talent
  Quote:
  
  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.
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distilabel_metadata.content: # Without L.R.,Long division,definition of derivative and eplison-delta prove, evaluate
  
  • Jan 13th 2010, 01:19 PM
  General
  Without L.R.,Long division,definition of derivative and eplison-delta proof, evaluate
  Without what I said, Evaluate :
  $\lim_{x \rightarrow 1} \frac{x^5-1}{x-1}$
  • Jan 13th 2010, 01:26 PM
  Jhevon
  Quote:
  
  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.
  • Jan 13th 2010, 01:27 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 01:32 PM
  Jhevon
  Quote:
  
  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
  • Jan 13th 2010, 01:33 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 01:41 PM
  Jhevon
  Quote:
  
  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?
  • Jan 13th 2010, 01:42 PM
  General
  Quote:
  
  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.
  • Jan 13th 2010, 02:21 PM
  Krizalid
  Quote:
  
  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. :D
  • Jan 13th 2010, 02:33 PM
  Jhevon
  Quote:
  
  Originally Posted by Krizalid
  I wouldn't recommend that if having $\frac{x^{n}-1}{x-1},$ you wouldn't expand. :D
  
  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?
  • Jan 13th 2010, 03:11 PM
  CaptainBlack
  Quote:
  
  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
  • Jan 14th 2010, 08:21 AM
  Jester
  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.
  • Jan 14th 2010, 07:24 PM
  I4talent
  Quote:
  
  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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a discussion from a mathematics forum where users are evaluating a limit problem without using certain methods (L'Hôpital's rule, long division, the definition of the derivative, and epsilon-delta proofs). The problem is to evaluate the limit:
  
  \[
  \lim_{x \rightarrow 1} \frac{x^5-1}{x-1}
  \]
  
  The discussion includes several users suggesting different methods to solve the problem. Here is a summary of the key points:
  
  1. **General** posed the problem and asked for solutions without using the restricted methods.
  2. **Jhevon** suggested factoring the numerator using the difference of powers formula: \(x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)\). This allows the \(x - 1\) terms to cancel, simplifying the limit to evaluating \(x^4 + x^3 + x^2 + x + 1\) at \(x = 1\).
  3. **General** noted that the factorization was derived using long division, which is not allowed.
  4. **Jhevon** provided a known fact about the factorization of \(a^n - b