Interpreting a function based on it's equation.

[Square1]

Ok so firstly, I'm not entirely sure where this thread should land so it's going in here lol.

So this question is inspired by doing preliminary algebra on a function in order to find a limit that initially results in an indeterminate form.

What does it say about what we get from the face value of a written equation, and what it actually is? For example,

lim x→1 of (x^2 - 1) / (x - 1)

makes the exact same looking graph on a calculator as (x-1) , but the first equation does not exist at 1, whereas the second one does! BLLAAAARRGGHHHHH!

 

[chiro]

Hey Square1 and welcome to the forums.

For this kind of problem you are indeed right in the confusion that arises and this is why there is a lot of 'funny things' that happen when you get the 0/0 form.

Having said that, there are rigorous proofs that allow us to evaluate limits such as these in certain ways and one of them is l'hospital's rule which allows you figure out these kind of situations.

Also what I am about to say is the most important fact of them all: Limits are used to figure out the behaviour of something when it gets infinitesimally close to a value but not at the value!

It seems crazy, but this is what calculus is based on. So by doing this we are not really considering a 0/0 case but a case where it gets really close to zero but not actually zero!

It is for this reason that we can do things like factor out terms since (x^2 - 1)/(x-1) = (x-1)(x+1)/(x-1) = (x+1) as long as x does not equal 1. But the limit doesn't talk about x actually equalling 1, but instead when it gets really really close to 1 (but still not actually 1!)

This is why we can do this kind of thing because we are not actually talking about evaluating a 0/0 but figuring out what it goes to as we approach it.

Then what you do is you take this result and add what is called continuity and what this will do is help you specify that the limit as well as the function value corresponding to information contained in the limit are the same. In other words if the derivative is analytic-continuous and the function is continuous then the information given by the limit represents the actual function itself.

I can go into this in more detail and clarify these things if you want, but the above should give you the idea of why we can do what we do.

 

[Mark44]

The graphs of y = (x² - 1)/(x - 1) and y = x + 1, except at x = 1, where the first function has a discontinuity at the point (1, 2). A single point has no width or length, so a graphing calculator won't show this discontinuity.

Note that you said that the first function has the same graph as y = x - 1, which isn't true.

 

[SteveL27]

Ok so firstly, I'm not entirely sure where this thread should land so it's going in here lol.

So this question is inspired by doing preliminary algebra on a function in order to find a limit that initially results in an indeterminate form.

What does it say about what we get from the face value of a written equation, and what it actually is? For example,

lim x→1 of (x^2 - 1) / (x - 1)

makes the exact same looking graph on a calculator as (x-1) , but the first equation does not exist at 1, whereas the second one does! BLLAAAARRGGHHHHH!

Well, the calculator failed basic algebra! Because dividing both the numerator and the denominator by x - 1 is only permissible when x doesn't equal 1. When x = 1 we can't do the division; and the function is not defined there.

 

[HallsofIvy]

The graph of x+ 1 is a straight line. The graph of (x^2-1)/(x- 1) is that same straight line except that it has a "hole" at (1, 2).

 

[Square1]

Note that you said that the first function has the same graph as y = x - 1, which isn't true.

Ok good point and you are right.

Also, yea as the second post says, its good to keep in mind the distinction between approaching a point and actually being on the point.

But hmm. How to put it...It's just kind of hitting me that the single process of simplification has the inherent property of taking an equation, and creating another one that is able to "pass the test" so to speak of finding that limit, and the answer applies to the first equation!

Ugh I feel as if I am not explaining this right. I'll just wait and see

 

[SteveL27]

Ok good point and you are right.

Also, yea as the second post says, its good to keep in mind the distinction between approaching a point and actually being on the point.

But hmm. How to put it...It's just kind of hitting me that the single process of simplification has the inherent property of taking an equation, and creating another one that is able to "pass the test" so to speak of finding that limit, and the answer applies to the first equation!

Ugh I feel as if I am not explaining this right. I'll just wait and see

You're right. It's a little counterintuitive to realize that f(x) = x is the not same function as f(x) = x^2/x. But as I indicated above, the equality x^2/x = x is only valid if x is different from 0. So f(x) = x^2/x is not defined at zero; although of course the limit as x->0 of x^2/x is in fact 0.

 

[Deveno]

the situation is a little clearer if one considers the function:

f(x) = x/x

(now we don't have any complicated algebra in the way).

well, as any fool can tell you, x/x = 1, right?

well, almost. 0/0 doesn't make any sense, because __ /0 is "bad" (undefined).

but, f(x) = x/x is perfectly reasonable everywhere EXCEPT 0, so if we made THIS function:

g(x) = x/x, x ≠ 0
g(0) = 1

the function g behaves the way we WANT f to behave.

in technical terms, g is CONTINUOUS, while f is NOT.

continuous functions, therefore, are functions that "make the sense they're supposed to". non-continuous functions, might do something "unexpected", so it might be best if we save studying those for later, after we understand continuous functions better.

of course, what I've done with the above discussion, is said basically:

if a function is continuous, certain kinds of troublesome problems go away.

well, that's all very well and good, but what, exactly, IS a continuous function (we've traded difficulty with functions, off for difficulty with some definition of a property a function might have)? and that's where the concept of a LIMIT comes in.

a limit acts like a number, or rather, acts like an ordinary "output" of a function, but it "extends the places a function can go". roughly speaking, a continuous function "equals its limits at all points".

now, we've made yet ANOTHER trade-off...to have any hope of knowing what continuous means, we need to know what a limit is. naively, a limit of a function, is what the output is near, when the input is near a given spot, say x = a. the exact definition of "near" requires strange quantifiers with epsilons and deltas, but in a loose sense, x near a means that the difference between x and a is small (close to 0, in a well-defined way).

so a continuous function is one for which f(x) is near f(a), when x is near a. this corresponds nicely to our naive idea that a continuos function of one variable, is something you can draw "without the pencil leaving the paper".

as you can see with f(x) = x/x, when we graph it, we have a microscopic gap at x = 0. such a small hole won't show up very well on a graph, but is disastrous if f represents something like an internet connection (the connection gets "broken" at x = 0, and that's that).

this is something to watch out for, for any function of the form:

f(x) = g(x)/h(x).

we might have h(a) = 0, for some value a, which might make f do some strange stuff near a.

 

[Square1]

*************************************************
Ok I'd like to have another go at this so I'll rephrase the question.
*************************************************

How is it that the process of factoring, expanding, or multiplying by a conjugate produces almost an identical function as the original.

A)What is so special about these processes that they can reproduce ALMOST THE EXACT SAME THING, ONLY WITH THE EXCEPTION OF A FEW POINTS?

B) so...(x^2-4)/(x+2) != (x-2) :O ??

 

[Mark44]

*************************************************
Ok I'd like to have another go at this so I'll rephrase the question.
*************************************************

How is it that the process of factoring, expanding, or multiplying by a conjugate produces almost an identical function as the original.

When you factor an expression, all you are doing is writing that expression in a new form that is identically equal to the original form. For example, x² + 2x + 1 = (x + 1)² is identically true (i.e., true for all values of the variable x).

When you expand an expression, you are just multiplying factors to get another expression that is identically equal to the first. In my equation above, you can expand (x + 1)² to get the expression shown on the left side.

You can always multiply an expression by 1 to get a new expression that has exactly the same value. If you multiply an expression by the conjugate, you will get a new expression with a different value. However, if you multiply the expression by the conjugate over itself, you are just multiplying by 1, so the new expression is identically equal to the one you started with.

A)What is so special about these processes that they can reproduce ALMOST THE EXACT SAME THING, ONLY WITH THE EXCEPTION OF A FEW POINTS?

B) so...(x^2-4)/(x+2) != (x-2) :O ??

Here we are factoring the left side to get
\frac{x^2 - 4}{x + 2} = \frac{(x-2)(x + 2)}{x + 2)} = (x - 2) \cdot \frac{x + 2}{x + 2}

The fraction on the right is equal to 1 except when x = -2, which also happens to be where the original rational expression on the left side is undefined.

 

[Square1]

Hmm I think I am coming closer. But still some q's..

\frac{x^2 - 4}{x + 2} = \frac{(x-2)(x + 2)}{x + 2)} = (x - 2) \cdot \frac{x + 2}{x + 2}

The fraction on the right is equal to 1 except when x = -2, which also happens to be where the original rational expression on the left side is undefined.

Hmm so both right and left side continue to be undefined at -2, and yet in one expression it is possible to calculate the value at -2!

I see the idea behind saying that (x+2) times (x-2)/(x-2) is like saying times 1, but (x-2)/(x-2) is only equal to one if x!=0. It is indeterminate form right?

 

[chiro]

Hmm I think I am coming closer. But still some q's..



Hmm so both right and left side continue to be undefined at -2, and yet in one expression it is possible to calculate the value at -2!

I see the idea behind saying that (x+2) times (x-2)/(x-2) is like saying times 1, but (x-2)/(x-2) is only equal to one if x!=0. It is indeterminate form right?

Its only equal to 1 when x != 2 since (x - 2) = 2 -2 = 0 which gives an indeterminant form expression.

 

[Square1]

Yea so that's kind of what I had said, and to repeat, we still end up calling (x-2)/(x-2) = 1 when we plug in -2, even though (x+2) times (x-2)/(x-2) is still, as said, undefined at -2.

 

[chiro]

Yea so that's kind of what I had said, and to repeat, we still end up calling (x-2)/(x-2) = 1 when we plug in -2, even though (x+2) times (x-2)/(x-2) is still, as said, undefined at -2.

This kind of thing is actually why we have to use limits in calculus.

In calculus we use limits to define things like the derivative and the reason we don't just plugin the 'h = 0' in place of 'h approaches 0' is for a similar reason in that we get an indeterminant form. We can't just let h = 0 because the same kind of situation that you have brought up.

But this kind of thing is a weird thing and you're definitely not the first person to ask about it.

 

[Square1]

haha yea. it definatly is occupying my thoughts at the moment :)

 

[Mark44]

Here we are factoring the left side to get
\frac{x^2 - 4}{x + 2} = \frac{(x-2)(x + 2)}{x + 2)} = (x - 2) \cdot \frac{x + 2}{x + 2}

The fraction on the right is equal to 1 except when x = -2, which also happens to be where the original rational expression on the left side is undefined.

Hmm I think I am coming closer. But still some q's..



Hmm so both right and left side continue to be undefined at -2, and yet in one expression it is possible to calculate the value at -2!

Not in what I wrote above. The three expressions in my equation are all undefined at x = -2.

I see the idea behind saying that (x+2) times (x-2)/(x-2) is like saying times 1, but (x-2)/(x-2) is only equal to one if x!=0. It is indeterminate form right?

You have switched some signs, and have another error. (x - 2) times (x + 2)/(x + 2) is the same as x - 2 times 1 if x != -2. The expression (x + 2)/(x + 2) has the indeterminate form [0/0].

 

[mjpam]

The limit of the rational function is indeterminate because the limit of the functionn depends on whether the numerator approaches zero or the denominator approaches zero.

 

[Mark44]

The limit of the rational function is indeterminate because the limit of the functionn depends on whether the numerator approaches zero or the denominator approaches zero.

If the rational function you refer to is (x² - 1)/(x - 1), the limit as x approaches 1 is NOT indeterminate.
\lim_{x \to 1}\frac{x^2 - 1}{x - 1} = 2

 

[mjpam]

If the rational function you refer to is (x² - 1)/(x - 1), the limit as x approaches 1 is NOT indeterminate.
\lim_{x \to 1}\frac{x^2 - 1}{x - 1} = 2

Uh...the form is indeterminate. That is why you can use L'Hopital's Rule to find the limit.

 

[Mark44]

Yes, the expression is one of several indeterminate forms, but the limit is 2, which is not indeterminate. In any case, a limit is never indeterminate.

 

[mjpam]

Yes, the expression is one of several indeterminate forms, but the limit is 2, which is not indeterminate. In any case, a limit is never indeterminate.

I think we're saying the same thing, or at least I'm trying to say the same thing as you.

My point is that the expression \frac{x^{2}-1}{x-1} is not itself indeterminate. Evaluated at any real-number value except x=1, it is single-valued and well-defined. However, at x=1, its denominator is zero, so its value is undefined.

The place where I am having some problem with terminology is here:

If we consider the above expression as a rational function R(x)=\frac{P(x)}{Q(x)} with P(x)=x^{2}-1 and Q(x)=x-1, we can use the heuristic I mentioned to understand why lim _{x \to 1}R(x) yields, at least in my words, an "indeterminate form":

I say that it is a heuristic because I am not letting P(x) and Q(x) vary simultaneously as they would when finding lim_{x \to 1}R(x), rather I am considering either P(1)=0 or Q(1)=0 and allowing the other function of x to vary as x \to 1 to show why, as I understand it, the form of the limit is indeterminate.

Basically if I let, P(x) vary, it yields a limit of the form lim_{x \to 1}=\frac{P(x)}{0}, which identically undefined everywhere except at x=1; whereas, if I let Q(x) vary, it yields a limit of the form lim_{x \to 0}=\frac{0}{Q(x)}, which identically 0 everywhere except at x=1. Because the limit has a different value depending on whether P(x) varies Q(x), the value of the limit in the form given is indeterminate. The limit does exist and is equal to the value already mentioned, but it is impossible to tell what its value is.

I'm not sure if I am expressing myself in the standard way, but that is how I understand the problem. I would appreciate any correction.

 

[Char. Limit]

I think we're saying the same thing, or at least I'm trying to say the same thing as you.

My point is that the expression \frac{x^{2}-1}{x-1} is not itself indeterminate. Evaluated at any real-number value except x=1, it is single-valued and well-defined. However, at x=1, its denominator is zero, so its value is undefined.

The place where I am having some problem with terminology is here:

If we consider the above expression as a rational function R(x)=\frac{P(x)}{Q(x)} with P(x)=x^{2}-1 and Q(x)=x-1, we can use the heuristic I mentioned to understand why lim _{x \to 1}R(x) yields, at least in my words, an "indeterminate form":

I say that it is a heuristic because I am not letting P(x) and Q(x) vary simultaneously as they would when finding lim_{x \to 1}R(x), rather I am considering either P(1)=0 or Q(1)=0 and allowing the other function of x to vary as x \to 1 to show why, as I understand it, the form of the limit is indeterminate.

Basically if I let, P(x) vary, it yields a limit of the form lim_{x \to 1}=\frac{P(x)}{0}, which identically undefined everywhere except at x=1; whereas, if I let Q(x) vary, it yields a limit of the form lim_{x \to 0}=\frac{0}{Q(x)}, which identically 0 everywhere except at x=1. Because the limit has a different value depending on whether P(x) varies Q(x), the value of the limit in the form given is indeterminate. The limit does exist and is equal to the value already mentioned, but it is impossible to tell what its value is.

I'm not sure if I am expressing myself in the standard way, but that is how I understand the problem. I would appreciate any correction.

That's definitely not the standard way. Your way is basically saying that the following expression is indeterminate:

\lim_{(x,y)\to(1,1)} \frac{y^2 - 1}{x-1}

Which is true, but much less meaningful. The standard sense of the limit has the two parts of the fraction tending toward 1 simultaneously. It's much less useful and much less meaningful otherwise.

 

[mjpam]

That's definitely not the standard way. Your way is basically saying that the following expression is indeterminate:

\lim_{(x,y)\to(1,1)} \frac{y^2 - 1}{x-1}

Which is true, but much less meaningful. The standard sense of the limit has the two parts of the fraction tending toward 1 simultaneously. It's much less useful and much less meaningful otherwise.

I was following Wikipedia's description of indeterminate forms. However, if we are going to discuss functions of two variable, my Calculus III professor gave somewhat of the same reason as to why the limits of multi-variable functions don't exist on higher dimensional real manifolds, and it's analogous to the explanation of why the limits of single-variable function don't exist for functions with "jumps" in them.

 

[Mark44]

I think we're saying the same thing, or at least I'm trying to say the same thing as you.

My point is that the expression \frac{x^{2}-1}{x-1} is not itself indeterminate. Evaluated at any real-number value except x=1, it is single-valued and well-defined. However, at x=1, its denominator is zero, so its value is undefined.

At x = 1, both the denominator and numerator are zero, which is what makes this expression indeterminate.

We can evaluate the limit to find that the discontinuity at x = 1 is removable, meaning that there is a "hole" at (1, 2).

 

[pwsnafu]

I was following Wikipedia's description of indeterminate forms.

Reread Wiki and I don't see how what is written up there gets you what you wrote.

However, if we are going to discuss functions of two variable, my Calculus III professor gave somewhat of the same reason as to why the limits of multi-variable functions don't exist on higher dimensional real manifolds, and it's analogous to the explanation of why the limits of single-variable function don't exist for functions with "jumps" in them.

No, what you are thinking about is path dependence. Single-variable jump functions can have a left limit and a right limit, if they do they won't be equal. That is a separate issue from what is being discussed in this thread.

Because the limit has a different value depending on whether P(x) or Q(x) varies, the value of the limit in the form given is indeterminate

No. Indeterminate does not mean that. To use Char. Limit's example,
\frac{y^2-1}{x-1}
is indeed in indeterminate as (x,y)->(1,1) because under the substitution (x,y)=(1,1) it becomes 0/0, and not because any dependence on paths.

You need to understand "undefined" "intermediate" and "limit does not exist" are three different terms. Path dependence uses the third not the second.

 

[Square1]

You need to understand "undefined" "intermediate" and "limit does not exist" are three different terms. Path dependence uses the third not the second.

Mind if I take the opportunity to review?

Undefined - a point for an equation does not exist. Found by dividing by zero, squaring a negative (other ways??)

Indeterminate - when a form is indeterminate, it takes the form of 0/0, infin/infin, infin - infin, 0 x infin, infin ^ 0, 0 ^ 0, c ^ infin (c is some constant)

Lim does not exist - right hand side and left hand side limits are not equal

that right?

 

[mjpam]

Reread Wiki and I don't see how what is written up there gets you what you wrote.

I recall the heuristics I described being on Wikipedia, but it is no longer there. Such is the curse of user-editable media.

No, what you are thinking about is path dependence.

Yes, it is, but I making an analogy to indeterminate forms, not saying that indeterminate forms and path dependence are one in the same.

Single-variable jump functions can have a left limit and a right limit, if they do they won't be equal.

I may be mistaken, but saying "the limit exists", without specifying the direction from which the limit is being taken, it usually taken to mean "the two-sided limit exists". That how I meant it, and, insofar as I assumed that that was the way it was going to be understood, functions with "jumps" in them do not have limits that exist at "jump" point because the right and left limits at those points are not equal.

That is a separate issue from what is being discussed in this thread.

I realize that, but I was making an analogy to path dependence, not stating its equivalence with indeterminate forms.

No. Indeterminate does not mean that. To use Char. Limit's example,
\frac{y^2-1}{x-1}
is indeed in indeterminate as (x,y)->(1,1) because under the substitution (x,y)=(1,1) it becomes 0/0, and not because any dependence on paths.

That is most certainly true. I, however, was trying to explain why the form of the limit under direct substitution is indeterminate, and not 0 or undefined.

You need to understand "undefined" "intermediate" and "limit does not exist" are three different terms. Path dependence uses the third not the second.

I do understand the difference, but similarly there is a difference between conflating two concepts and analogizing them

 

[Mark44]

Mind if I take the opportunity to review?

Undefined - a point for an equation does not exist. Found by dividing by zero, squaring a negative (other ways??)

A value for which an expression does not exist. It makes no sense to say that an equation doesn't exist.

Possible ways an expression can fail to exist: division by zero; taking the square root (or an even root) of a negative number - there's no problem with squaring a negative number; attempting to evaluate a function at a value outside its domain (e.g., ln(0)), and so on.

Indeterminate - when a form is indeterminate, it takes the form of 0/0, infin/infin, infin - infin, 0 x infin, infin ^ 0, 0 ^ 0, c ^ infin (c is some constant)

Also [-∞/∞], but c^∞ is not indeterminate.

Lim does not exist - right hand side and left hand side limits are not equal

Which includes the possibility that either fails to exist. For example,
\lim_{x \to 0} \sqrt{x} \text{ does not exist}

that right?

 

[Square1]

Wow it's almost a year since I've first seen this material and I am running back into it. Geeze I remember vividly how much this made my head spin :D lol - it was one of the most confusing things for me in calculus along with l'hospitals rule etc. Now it so much more clear reading the textbook and understanding it all. And I remember that this thread provided lot's of help too at the time and I made reference to it a few times afterwards. So thanks a lot everyone. All the responses are top notch!