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Interpreting a function based on it's equation.

by Square1
Tags: based, equation, function, interpreting
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mjpam
#19
Feb20-12, 06:20 PM
P: 76
Quote Quote by Mark44 View Post
If the rational function you refer to is (x2 - 1)/(x - 1), the limit as x approaches 1 is NOT indeterminate.
[tex]\lim_{x \to 1}\frac{x^2 - 1}{x - 1} = 2[/tex]
Uh....the form is indeterminate. That is why you can use L'Hopital's Rule to find the limit.
Mark44
#20
Feb20-12, 06:35 PM
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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
#21
Feb20-12, 07:44 PM
P: 76
Quote Quote by Mark44 View Post
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 [itex]\frac{x^{2}-1}{x-1}[/itex] is not itself indeterminate. Evaluated at any real-number value except [itex]x=1[/itex], it is single-valued and well-defined. However, at [itex]x=1[/itex], 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 [itex]R(x)=\frac{P(x)}{Q(x)}[/itex] with [itex]P(x)=x^{2}-1[/itex] and [itex]Q(x)=x-1[/itex], we can use the heuristic I mentioned to understand why [itex]lim _{x \to 1}R(x)[/itex] yields, at least in my words, an "indeterminate form":

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

Basically if I let, [itex]P(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 1}=\frac{P(x)}{0}[/itex], which identically undefined everywhere except at [itex]x=1[/itex]; whereas, if I let [itex]Q(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 0}=\frac{0}{Q(x)}[/itex], which identically 0 everywhere except at [itex]x=1[/itex]. Because the limit has a different value depending on whether [itex]P(x)[/itex] varies [itex]Q(x)[/itex], 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
#22
Feb20-12, 07:53 PM
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Quote Quote by mjpam View Post
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 [itex]\frac{x^{2}-1}{x-1}[/itex] is not itself indeterminate. Evaluated at any real-number value except [itex]x=1[/itex], it is single-valued and well-defined. However, at [itex]x=1[/itex], 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 [itex]R(x)=\frac{P(x)}{Q(x)}[/itex] with [itex]P(x)=x^{2}-1[/itex] and [itex]Q(x)=x-1[/itex], we can use the heuristic I mentioned to understand why [itex]lim _{x \to 1}R(x)[/itex] yields, at least in my words, an "indeterminate form":

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

Basically if I let, [itex]P(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 1}=\frac{P(x)}{0}[/itex], which identically undefined everywhere except at [itex]x=1[/itex]; whereas, if I let [itex]Q(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 0}=\frac{0}{Q(x)}[/itex], which identically 0 everywhere except at [itex]x=1[/itex]. Because the limit has a different value depending on whether [itex]P(x)[/itex] varies [itex]Q(x)[/itex], 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:

[tex]\lim_{(x,y)\to(1,1)} \frac{y^2 - 1}{x-1}[/tex]

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
#23
Feb20-12, 08:17 PM
P: 76
Quote Quote by Char. Limit View Post
That's definitely not the standard way. Your way is basically saying that the following expression is indeterminate:

[tex]\lim_{(x,y)\to(1,1)} \frac{y^2 - 1}{x-1}[/tex]

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
#24
Feb20-12, 08:37 PM
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Quote Quote by mjpam View Post
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 [itex]\frac{x^{2}-1}{x-1}[/itex] is not itself indeterminate. Evaluated at any real-number value except [itex]x=1[/itex], it is single-valued and well-defined. However, at [itex]x=1[/itex], 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
#25
Feb20-12, 08:46 PM
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P: 837
Quote Quote by mjpam View Post
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,
[itex]\frac{y^2-1}{x-1}[/itex]
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
#26
Feb21-12, 12:26 AM
P: 115
Quote Quote by pwsnafu View Post
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
#27
Feb21-12, 02:15 PM
P: 76
Quote Quote by pwsnafu View Post
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.

Quote Quote by pwsnafu View Post
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.

Quote Quote by pwsnafu View Post
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.

Quote Quote by pwsnafu View Post
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.

Quote Quote by pwsnafu View Post
No. Indeterminate does not mean that. To use Char. Limit's example,
[itex]\frac{y^2-1}{x-1}[/itex]
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.

Quote Quote by pwsnafu View Post
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
#28
Feb21-12, 03:59 PM
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Quote Quote by Square1 View Post
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.
Quote Quote by Square1 View Post

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.
Quote Quote by Square1 View Post

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,
[tex]\lim_{x \to 0} \sqrt{x} \text{ does not exist}[/tex]
Quote Quote by Square1 View Post
that right?
Square1
#29
Sep20-12, 09:13 PM
P: 115
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!


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