How to think about x/x, or (x-1)/(x-1) etc.

[kmm]

In solving physics problems, I have often done some type of simplifying where I eliminated an x in the numerator and denominator, or eliminated some other terms. For example, maybe I have something like $\frac {x} {x^2 + x}$ and I simplify this to $\frac {1} {x+1}$. Or I have something like $\frac {x-1} {x^2-2x+1} \to \frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$. Typically, this is something I've taken for granted and done without much thought. But it dawned on me that in the first example, the original function is undefined at x=0, and the simplified version is one, at x=0 but has asymptotic behavior around x=-1. Also to get to this simplified version, by factoring out an x in the numerator and denominator, I had to assume $\frac {x} {x} =1$. But $\frac {x} {x}$ is undefined at x=0, so how can I simply assume it is 1? For the second example, $\frac {x-1} {x^2-2x+1}$ is undefined at x=1, but $\frac {1} {x-1}$ goes to infinity at x=1. And to go from $\frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$, I again had to assume that $\frac {x-1} {x-1} = 1$, but this is undefined at x=1. I'm not sure how I am supposed to be thinking about all of this. Is there some sort of redefining of process of these functions for these situations? I'm aware of L'Hostpitals Rule for indeterminate forms, but it has never occurred to me that I should use it in these situations where I'm simplifying a basic function.

 

[jedishrfu]

In the first case and it’s simplifying case as x approaches zero the function approaches 1

In physics we would restrict the boundary conditions to not allow x to be zero in these cases and then everything works as expected.

When I was an undergrad physics major, we were taught that math is the language of physics and a tool of physics but we didn’t worry so much about the rigor only about the boundarymconditions to avoid infinities or undefined values.

 

[fresh_42]

I'm not sure how I am supposed to be thinking about all of this. Is there some sort of redefining of process of these functions for these situations? I'm aware of L'Hostpitals Rule for indeterminate forms, but it has never occurred to me that I should use it in these situations where I'm simplifying a basic function.

There are different kinds of such points, i.e. singularities, and how to deal with them or how to think of them depends on what you want to do. The Wikipedia link will give you some insights.

In physics we would restrict the boundary conditions to not allow x to be zero in these cases and then everything works as expected.

That's why mathematicians categorized them and founded measure theory. And without much effort silently changed the physicists' way of integration form Riemann (Archimedes) to Lebesgue (QM)

 

[Mark44]

In solving physics problems, I have often done some type of simplifying where I eliminated an x in the numerator and denominator, or eliminated some other terms. For example, maybe I have something like $\frac {x} {x^2 + x}$ and I simplify this to $\frac {1} {x+1}$. Or I have something like $\frac {x-1} {x^2-2x+1} \to \frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$. Typically, this is something I've taken for granted and done without much thought. But it dawned on me that in the first example, the original function is undefined at x=0, and the simplified version is one, at x=0 but has asymptotic behavior around x=-1. Also to get to this simplified version, by factoring out an x in the numerator and denominator, I had to assume $\frac {x} {x} =1$. But $\frac {x} {x}$ is undefined at x=0, so how can I simply assume it is 1?

Because for every value of x except 0, $\frac x x$ is exactly equal to 1. There is a singularity, in the form of a point discontinuity, at (0, 1). Except for this point the graph of $y = \frac x x$ is identical to the graph of $y = 1$.

For the second example, $\frac {x-1} {x^2-2x+1}$ is undefined at x=1, but $\frac {1} {x-1}$ goes to infinity at x=1.

That's not quite right. The left-side limit is $-\infty$ and the right-side limit is $\infty$.

And to go from $\frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$, I again had to assume that $\frac {x-1} {x-1} = 1$, but this is undefined at x=1. I'm not sure how I am supposed to be thinking about all of this.

The same as in your example of $\frac x x$. The graph of $y = \frac{x - 1}{x - 1}$ is identical to the graph of $y = 1$ except for a point discontinuity at (1, 1).

Is there some sort of redefining of process of these functions for these situations? I'm aware of L'Hostpitals Rule for indeterminate forms, but it has never occurred to me that I should use it in these situations where I'm simplifying a basic function.

When you simplify something like $\frac x {x^2 + x}$ to $\frac 1 {x + 1}$ you should explicitly state that x cannot be 0. It's not necessary to state that x can't be -1, since it is clearly not in the domain of either expression.

In another of your examples, simplifying $\frac{x - 1}{(x - 1)^2}$ to $\frac 1 {x - 1}$, it's not necessary to state that $x \ne 1$, because it's obvious in both expressions that x can't be 1.

 

[kmm]

That's not quite right. The left-side limit is $-\infty$ and the right-side limit is $\infty$.

That's right, thanks for pointing that out.

When you simplify something like $\frac x {x^2 + x}$ to $\frac 1 {x + 1}$ you should explicitly state that x cannot be 0. It's not necessary to state that x can't be -1, since it is clearly not in the domain of either expression.

In another of your examples, simplifying $\frac{x - 1}{(x - 1)^2}$ to $\frac 1 {x - 1}$, it's not necessary to state that $x \ne 1$, because it's obvious in both expressions that x can't be 1.

This is a helpful guideline.

In physics we would restrict the boundary conditions to not allow x to be zero in these cases and then everything works as expected.

I had suspected this would be the case.

There are different kinds of such points, i.e. singularities, and how to deal with them or how to think of them depends on what you want to do. The Wikipedia link will give you some insights.

Thanks for the link. I'm going to have to take some time to digest this.

 

[FactChecker]

In applying these types of equations to physics problems, you are wise to be aware that there are values of x where your calculations may be questionable. You should keep track of those values and see if they really do present a problem in the physical situation. I think that you will usually find that they are not a problem because the physical situation behaves continuously at those points and that your simplified equations are applicable there.

 

[bhobba]

I mentioned such things when I did my math degree. The out they told me was you use the extended reals:
https://en.wikipedia.org/wiki/Extended_real_number_line

That seemed to satisfy me at the time. Haven't thought about it since. I think these days the rigorous way would be using the hyper-reals - but you do not want to investigate that - it is hard - but the results are reasonably intuitive:
https://en.wikipedia.org/wiki/Hyperreal_number

Thanks
Bill

 

[PeroK]

In solving physics problems, I have often done some type of simplifying where I eliminated an x in the numerator and denominator, or eliminated some other terms. For example, maybe I have something like $\frac {x} {x^2 + x}$ and I simplify this to $\frac {1} {x+1}$. Or I have something like $\frac {x-1} {x^2-2x+1} \to \frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$. Typically, this is something I've taken for granted and done without much thought. But it dawned on me that in the first example, the original function is undefined at x=0, and the simplified version is one, at x=0 but has asymptotic behavior around x=-1. Also to get to this simplified version, by factoring out an x in the numerator and denominator, I had to assume $\frac {x} {x} =1$. But $\frac {x} {x}$ is undefined at x=0, so how can I simply assume it is 1? For the second example, $\frac {x-1} {x^2-2x+1}$ is undefined at x=1, but $\frac {1} {x-1}$ goes to infinity at x=1. And to go from $\frac {x-1} {(x-1)^2} \to \frac {1} {x-1}$, I again had to assume that $\frac {x-1} {x-1} = 1$, but this is undefined at x=1. I'm not sure how I am supposed to be thinking about all of this. Is there some sort of redefining of process of these functions for these situations? I'm aware of L'Hostpitals Rule for indeterminate forms, but it has never occurred to me that I should use it in these situations where I'm simplifying a basic function.

When you eliminate a common factor from a fraction you should get into the habit of noting the value(s) for which the process does not hold. E.g.

$y = \frac{x^2}{x}$

$y = x \ (x \ne 0)$

You should do this automatically as good maths technique.

 

[hilbert2]

This is called a removable discontinuity, which can be handled by redefining the function at a single point (unlike a function $f(x)=1/x$ where there's no way to make it continuous at $x=0$).

 

[jedishrfu]

I mentioned such things when I did my math degree. The out they told me was you use the extended reals:
https://en.wikipedia.org/wiki/Extended_real_number_line

That seemed to satisfy me at the time. Haven't thought about it since. I think these days the rigorous way would be using the hyper-reals - but you do not want to investigate that - it is hard - but the results are reasonably intuitive:
https://en.wikipedia.org/wiki/Hyperreal_number

Thanks
Bill

With respect to the hyper-reals, they were created to eliminate the traditional limit pedagogy concept as a basis for calculus rigor that caused much pain to students of calculus with a number system that implicitly supported it. The key was the transfer principle that relates properties of hyper-reals to reals allowing one to use algebra to reduce limit expressions without a concern for zero denominators. Basically it allowed for algebraic math with infinitesimal quantities.

As an example, deriving the derivative of $y=sin(x)$ one could write:

$dy/dx = (sin(x+dx) - sin(x))/dx$

$dy/dx = (sin(x)cos(dx)+cos(x)sin(dx) - sin(x)) / dx$

then realizing that $cos(dx) = 1$ as dx becomes 0 and $sin(dx) = dx$ then we get

$dy/dx = (sin(x) + cos(x)*dx - sin(x)) / dx$

$dy/dx = ( cos(x)*dx ) / dx$ realizing $sin(x) - sin(x) = 0$

$dy/dx = cos(x)$ realizing $dx / dx = 1$ since $dx$ is an infinitesimal ie not zero you are allowed to simplify it algebraically

You can read more about this in Keisler'a Calculus book:

https://www.math.wisc.edu/~keisler/calc.html