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

[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!

 

[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.

 

[nobahar]

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.

 

[hunt_mat]

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

 

[nobahar]

you want
<br /> \lim_{x\rightarrow 1}\frac{x^{10}-1}{x^{2}-1}<br />
The top factorises as:
<br /> x^{10}-1=(x-1)(x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)<br />
The bottom factorises as:
<br /> x^{2}-1=(x+1)(x-1)<br />
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.

\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 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.

 

[hunt_mat]

The bottom factorises as (x-1)(x-3) and you can compute the factorisation of the top by writing:
<br /> x^{4}+x^{3}-2=(x-1)(x^{3}+ax^{2}+bx+c)<br />
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.

 

[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.

 

[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.

 

[eumyang]

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 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

 

[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.

 

[vela]

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

 

[nobahar]

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!

 

[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}))

 

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

 

[nobahar]

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

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.

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}))

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

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:
<br /> p(x)=(x-a)q(x)<br />
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.