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

In summary, the conversation discusses finding the limit of a function using l'Hopital's Rule and factoring polynomials. The conversation also touches on the concept of prime polynomials and the fundamental theorem of algebra.
  • #1
nobahar
497
2
Hello!
This is another conceptual understanding question. Any help is appreciated.
I want to solve the following:

The limit as x -> 1 of:
[tex]\frac{x^{10}-1}{x^2-1}[/tex]

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:
Physics news on Phys.org
  • #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.
 
  • #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.
 
  • #4
you want
[tex]
\lim_{x\rightarrow 1}\frac{x^{10}-1}{x^{2}-1}
[/tex]
The top factorises as:
[tex]
x^{10}-1=(x-1)(x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)
[/tex]
The bottom factorises as:
[tex]
x^{2}-1=(x+1)(x-1)
[/tex]
Does this help?
 
  • #5
hunt_mat said:
you want
[tex]
\lim_{x\rightarrow 1}\frac{x^{10}-1}{x^{2}-1}
[/tex]
The top factorises as:
[tex]
x^{10}-1=(x-1)(x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)
[/tex]
The bottom factorises as:
[tex]
x^{2}-1=(x+1)(x-1)
[/tex]
Does this help?
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 realized that when the function yields 0 at x =a, x-a will be a factor. I apologise, I realize 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:
  • #6
The bottom factorises as [tex](x-1)(x-3)[/tex] and you can compute the factorisation of the top by writing:
[tex]
x^{4}+x^{3}-2=(x-1)(x^{3}+ax^{2}+bx+c)
[/tex]
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.
 
  • #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.
 
  • #8
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.
 
  • #9
nobahar said:
EDIT: I have realized that when the function yields 0 at x =a, x-a will be a factor. I apologise, I realize 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))...
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]


69
 
  • #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.
 
  • #11
I wouldn't say it's self-evident, but it is a consequence of the fundamental theorem of algebra.
 
  • #12
Thanks vela.
I'll look into 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:
  • #13
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:
  • #14
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:
[tex]
p(x)=(x-a)q(x)
[/tex]
You can convice youself of this by looking at a few cases yourself (cubics and quadratics)
 
  • #15
HallsofIvy said:
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).

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.

HallsofIvy said:
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]

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

hunt_mat said:
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:
[tex]
p(x)=(x-a)q(x)
[/tex]
You can convice youself of this by looking at a few cases yourself (cubics and quadratics)

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:

1. What is the limit of the fraction as x approaches 1?

The limit of the fraction as x approaches 1 is undefined. This is because when x is equal to 1, the denominator becomes 0, which is an undefined value in mathematics.

2. Can the limit of the fraction be evaluated by plugging in x=1?

No, the limit cannot be evaluated by plugging in x=1. This is because plugging in 1 for x would result in a division by 0, which is undefined.

3. What is the significance of the numerator and denominator in this limit?

The numerator and denominator play important roles in determining the behavior of the limit. For this specific fraction, as x gets closer to 1, the numerator approaches 0 while the denominator approaches 0 as well. This results in an indeterminate form known as 0/0, which cannot be directly evaluated.

4. How can the limit of this fraction be evaluated?

The limit of this fraction can be evaluated using algebraic manipulation or by applying L'Hopital's rule. Algebraic manipulation involves factoring out the common factor of (x-1) from both the numerator and denominator, which results in a simpler expression. L'Hopital's rule states that for a fraction where both the numerator and denominator approach 0, the limit can be found by taking the derivative of the numerator and denominator and then evaluating the limit again.

5. What is the general rule for finding the limit of a fraction?

The general rule for finding the limit of a fraction is to evaluate the limit of the numerator and denominator separately, and then divide the resulting limits. If the denominator is approaching 0, the limit is undefined. If the numerator and denominator both approach 0, further manipulation or the application of L'Hopital's rule may be needed to determine the limit.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
790
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
588
  • Calculus and Beyond Homework Help
Replies
8
Views
897
  • Calculus and Beyond Homework Help
Replies
10
Views
792
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
465
  • Calculus and Beyond Homework Help
Replies
6
Views
505
  • Calculus and Beyond Homework Help
Replies
12
Views
697
  • Calculus and Beyond Homework Help
Replies
17
Views
485
Back
Top