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Homework Help: Limit of a fraction (x^10-1)/(x^2-1)

  1. Aug 14, 2010 #1
    This is another conceptual understanding question. Any help is appreciated.
    I want to solve the following:

    The limit as x -> 1 of:

    So I divide the numerator and denominator by x-1 and then find the derivative of them separately with respect to x. Then take the ratio of the two derivatives.
    My question is: why divide by x-1?
    Is it so that it fits the form of the following derivative?:
    The limit as x -> x0 of:

    [tex]\frac{f(x_0 + dx) - f(x_0)}{x-x_0}[/tex]

    [tex]\frac{x^{10}-1 - (1^10-1)}{x-1}[/tex]

    So I'm assuming x0= 1 and x(^10) -1 is a f(x)?
    Thanks in advance. I can elaborate if it's not clear, I was more worried about the Latex not working!
    Last edited: Aug 14, 2010
  2. jcsd
  3. Aug 14, 2010 #2
    The easy way to find the limit is to simply apply l'Hopital's Rule to the fraction as it stands.

    If you follow the first procedure you outlined above-- first divide by x-1, then apply l'Hopitals's Rule-- you are making a mistake. That is because after dividing by x-1 the fraction no longer has the form 0/0, and l'Hopital's Rule does not apply.
  4. Aug 14, 2010 #3
    Thanks for the reply awkward.
    I'm sorry, I don't understand. Just as a precedent, this is intended to be 'pre-l'hospital'.
    If I divide by x-1 for both the numerator and denomintor, the limit as x tends to 1 is still 0/0; isn't it? So all I'm doing is evaluating the derivative with respect to x at x=1?
    I can't see x-1 being a common 'factor' of any sort to both x^10 -1 or x^2 -1.
    Any further help appreciated.
  5. Aug 14, 2010 #4


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    you want
    \lim_{x\rightarrow 1}\frac{x^{10}-1}{x^{2}-1}
    The top factorises as:
    The bottom factorises as:
    Does this help?
  6. Aug 15, 2010 #5
    Thanks hunt_mat.
    Blimey, I don't know how I missed that...

    EDIT: Okay, just editing this post because I made a mistake, I believed the following did not have a common factor of (x-1), but it does.

    [tex]\frac{x^4 + x^3 -2}{x^2 -4x +3}[/tex]

    At x=1 this is an indeterminate. But (I think) I can evaluate it by diving both the numerator and the denominator by x-1, as it fits the difference quotient where both the limits as x tends to 1 of both r(x)/(x-1) and q(x)/(x-1) are r'(1) and q'(1) - or that it is merely the derivatives of the functions with respect to x evaluated at x=1.

    EDIT: I have realised that when the function yields 0 at x =a, x-a will be a factor. I apologise, I realise this is basic stuff but I completely forgot this point! I NOW THINK THIS IS NOT TRUE!!! For example x^2 -2x -4 has a solution 1+sqrt(5), but it does not factor to give (x-(1+sqrt(5))...
    Is the reasoning behind performing the division by a common factor and taking the limit with respect to x separately in the numerator and denominator correct? That it is the f' at x=1?
    Again, many thanks.
    Last edited: Aug 15, 2010
  7. Aug 15, 2010 #6


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    The bottom factorises as [tex](x-1)(x-3)[/tex] and you can compute the factorisation of the top by writing:
    This is the usual method for problems like these. we know that the top has a factor (x-1) because 1^4+1^3-2=0.
  8. Aug 16, 2010 #7
    Thanks again hunt mat.
    If x=a yields a solution such that the polynomial p(a) = 0. Does it always follow that (x-a) is a factor? I am thinking in terms of what I have found are called prime polynomials. These have solutions in the real numbers but do not factor in such a way. Is it possible for two prime polynomials to have the same solution (that at x = a, both polynomials = 0)?
    Any further help appreciated.
  9. Aug 16, 2010 #8


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    It is always true that if [tex]p(a)=0[/tex] for a polynomial then [tex]x-a[/tex] will be a factor of that polynomial. I don't know what a prime polynomial is but over the reals that all factors are either of the form [tex]x-\alpha[/tex] or [tex]ax^{2}+bx+c[/tex]. The [tex]\alpha[/tex] may be irrational.
  10. Aug 16, 2010 #9


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    Yes, it does. Not sure why you think otherwise. Now true, the quadratic [tex]x^2 - 2x - 4[/tex] does not factor over the integers, but it does factor over the irrationals. The complete linear factorization would be
    [tex]x^2 - 2x - 4 = [x - (1 + \sqrt{5})][x - (1 - \sqrt{5})][/tex]

  11. Aug 16, 2010 #10
    Thanks hunt mat and eumyang for clearing that up for me.
    I can see that if (x-a) is a factor of a polynomial, then it is self evident that a is a root of the polynomial. However, should it be self-evident that the polynomial will always be factorable in such a way that (x-a) arises? How do I disregard the possibility that the polynomnial is not factorable (apologies, this is what I meant by a prime polynomial), or if it can be 'tidied up' (sorry for non-mathematical lingo!) in such a way that the factor (x-a) arises?
    I tried to find an explanation on the internet. Wiki merely states that (x-r) will be a factor if p(r)=0.

    N.B. I won't attempt any more examples!

    Many thanks for the help.
    Any further help appreciated.
  12. Aug 16, 2010 #11


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    I wouldn't say it's self-evident, but it is a consequence of the fundamental theorem of algebra.
  13. Aug 16, 2010 #12
    Thanks vela.
    I'll look in to the fundamental theorem. Hopefully, there's a simple proof. I'm also somewhat relieved it's not self-evident!

    EDIT: I looked, there's a proof, but apparently its obscenely difficult!
    Last edited: Aug 16, 2010
  14. Aug 17, 2010 #13


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    Not all that difficult. By the "Euclidean algorithm", If P(x) is a polynomial and a any number then there exists some polynomial Q(x), of degree one less than P(x), and number r such that P(x)= Q(x)(x- a)+ r (Q is the "quotient" and r is the "remainder"). Of course, setting x= a in that gives P(a)= r. If, in fact, P(a)= 0, then we must have P(x)= Q(x)(x- a)+ 0= Q(x)(x- a) so that x- a is a factor of P(x).

    By the way, the simplest way to find the factorization of [itex]x^2- 2x- 4[/itex] given above is to "complete the square". -2/2= -1 and (-1)2= 1 so we have [itex]x^2- 2x- 4= x^2- 2x+ 1- 1- 4= (x- 1)^2- 5[/itex]. That is a "difference of squares", [itex](x- 1)^2- (\sqrt{5})^2[/itex] so that
    [tex]x^2- 2x- 4= (x-1-\sqrt{5})(x-1+ \sqrt{5})= (x- (1+\sqrt{5}))(x- (1- \sqrt{5}))[/tex]
    Last edited by a moderator: Aug 17, 2010
  15. Aug 17, 2010 #14


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    One thing you can say is that suppose that a isn't a zero of a polynomial p(x), then it is impossible to find a factorisation such that:
    You can convice youself of this by looking at a few cases yourself (cubics and quadratics)
  16. Aug 18, 2010 #15
    Thanks Halls, my previous comment was based on this source from the web:
    http://mathdl.maa.org/images/upload_library/22/Polya/07468342.di020748.02p0019l.pdf [Broken]

    I don't know if I fully understand how it follows from the euclidean algorithm (although I haven't previously heard of it). I'll look into it further.
    In the meantime, is the following line of reasoning anywhere near plausible? To get p(x), y(x)*q(x) + remainder is always a legitimate way because the remainder will 'compensate', or 'make up the difference', for the product. Therefore, if y(x)=(x-a), then it follows, as Halls explained, that the remainder will equal zero, and so q(x)(x-a) will always be possible?
    Thanks for any help on this.

    Thanks Halls, I've yet to study polynomials. Can anyone recommend which to pursue first, matrices or polynomials?

    Thanks hunt_mat. But following my reasoning above, if a is not a root, (x-a)q(x) + r is still possible, isn't it?

    Thanks again for all the help, it's much appreciated.
    Last edited by a moderator: May 4, 2017
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