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

1. Aug 14, 2010

nobahar

Hello!
This is another conceptual understanding question. Any help is appreciated.
I want to solve the following:

The limit as x -> 1 of:
$$\frac{x^{10}-1}{x^2-1}$$

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:

$$\frac{f(x_0 + dx) - f(x_0)}{x-x_0}$$

$$\frac{x^{10}-1 - (1^10-1)}{x-1}$$

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. Aug 14, 2010

awkward

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. Aug 14, 2010

nobahar

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. Aug 14, 2010

hunt_mat

you want
$$\lim_{x\rightarrow 1}\frac{x^{10}-1}{x^{2}-1}$$
The top factorises as:
$$x^{10}-1=(x-1)(x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)$$
The bottom factorises as:
$$x^{2}-1=(x+1)(x-1)$$
Does this help?

5. Aug 15, 2010

nobahar

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.

$$\frac{x^4 + x^3 -2}{x^2 -4x +3}$$

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
6. Aug 15, 2010

hunt_mat

The bottom factorises as $$(x-1)(x-3)$$ and you can compute the factorisation of the top by writing:
$$x^{4}+x^{3}-2=(x-1)(x^{3}+ax^{2}+bx+c)$$
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. Aug 16, 2010

nobahar

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. Aug 16, 2010

hunt_mat

It is always true that if $$p(a)=0$$ for a polynomial then $$x-a$$ 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 $$x-\alpha$$ or $$ax^{2}+bx+c$$. The $$\alpha$$ may be irrational.

9. Aug 16, 2010

eumyang

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

69

10. Aug 16, 2010

nobahar

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. Aug 16, 2010

vela

Staff Emeritus
I wouldn't say it's self-evident, but it is a consequence of the fundamental theorem of algebra.

12. Aug 16, 2010

nobahar

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
13. Aug 17, 2010

HallsofIvy

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 $x^2- 2x- 4$ given above is to "complete the square". -2/2= -1 and (-1)2= 1 so we have $x^2- 2x- 4= x^2- 2x+ 1- 1- 4= (x- 1)^2- 5$. That is a "difference of squares", $(x- 1)^2- (\sqrt{5})^2$ so that
$$x^2- 2x- 4= (x-1-\sqrt{5})(x-1+ \sqrt{5})= (x- (1+\sqrt{5}))(x- (1- \sqrt{5}))$$

Last edited by a moderator: Aug 17, 2010
14. Aug 17, 2010

hunt_mat

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:
$$p(x)=(x-a)q(x)$$
You can convice youself of this by looking at a few cases yourself (cubics and quadratics)

15. Aug 18, 2010

nobahar

Thanks Halls, my previous comment was based on this source from the web: