.999 does not equal 1 because

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  • #26
Redbelly98
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I disagree with [itex]1^{+\infty}[/itex] is indeterminate.
If my initial equations are correct then [itex]1^{+\infty}[/itex] is bounded by e, 1/e.
Here's a counterexample to that statement:

limn→∞ [(1+10-n)2·10n] = e2

Replace the "2" with larger numbers, and it's evident that 1 is not only indeterminant but unbounded as well.
 
  • #27
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That's really akin to asking whether the geometrical line, infinite sets, ordered pairs, functions or relations really exist. Do ideas exist?
Yes, but then it seems to me that such ideas are necessarily ambiguous. I think there are theorems about this that make such statements more rigorous, perhaps Hurkyl can explain that better (I don't know much about the foundations of mathematics).
 
  • #28
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Here's a counterexample to that statement:

limn→∞ [(1+10-n)2·10n] = e2

Replace the "2" with larger numbers, and it's evident that 1 is not only indeterminant but unbounded as well.
I disagree, if you do a binomial expansion of
limn→∞ [(1+10-n)2·10n] = e2
The first term is [itex]1^{+\infty}[/itex] which you say is indeterminate, yet you seem happy to have a definite answer to this expression. Extrapolating n to increasing numbers does indicate convergence to a definite value.
Further, at least for finite n, [tex]\(1 + 10^{-n}[/tex] >1. Your equation uses [tex]\(1 + 10^{-n}[/tex] in place of 1, and demonstrates that [tex]\(1 + 10^{-n}[/tex] >1 for infinite n at least for this expression.
 
  • #29
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I disagree, if you do a binomial expansion of
limn→∞ [(1+10-n)2·10n] = e2
The first term is [itex]1^{+\infty}[/itex] which you say is indeterminate, yet you seem happy to have a definite answer to this expression. Extrapolating n to increasing numbers does indicate convergence to a definite value.
Redbelly98 is not saying that [itex]1^{\infty}[/itex] has a definite value. To the contrary, he is demonstrating an expression of this indeterminate form that is a counterexample to your assertion that [itex]1^{\infty}[/itex] is bounded by 1/e and e.

All of the indeterminate forms, of which [[itex]1^{\infty}[/itex]] is just one, are called indeterminate because you can't a priori determine a value for them, or even that they have a value.

Some of the other indeterminate forms are [0/0], [[itex]\infty/\infty[/itex]], and [[itex]\infty - \infty[/itex]].
 
  • #30
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Redbelly98 is not saying that [itex]1^{\infty}[/itex] has a definite value. To the contrary, he is demonstrating an expression of this indeterminate form that is a counterexample to your assertion that [itex]1^{\infty}[/itex] is bounded by 1/e and e.

All of the indeterminate forms, of which [[itex]1^{\infty}[/itex]] is just one, are called indeterminate because you can't a priori determine a value for them, or even that they have a value.

Some of the other indeterminate forms are [0/0], [[itex]\infty/\infty[/itex]], and [[itex]\infty - \infty[/itex]].
You are saying that [itex]\lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n}) [/itex]
But I am saying that the limit is never actually reached, the expressions fall short by an infinitesimal amount.
If you treat the above variations of 1 as interchangeable then I agree that [itex]1^{\infty}[/itex] is indeterminate. However my earlier expressions give different results for these different versions of 1.
Please could someone explain this to me? What is the reason for this convention, and why do my expressions not disprove it?
Also, how can my expressions have a definite value if [itex]1^{\infty}[/itex] does not, as my expression is just a variation of [itex]1^{\infty}[/itex]?
 
  • #31
Hurkyl
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You are saying that [itex]\lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n}) [/itex]
But I am saying that the limit is never actually reached, the expressions fall short by an infinitesimal amount.
The following statement is true:
For any real number n, 1 + 10-n does not equal 1​
The following statement is false:
The limit of 1 + 10-n as n approaches infinity does not equal 1​
 
  • #32
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Without trying to make myself sound like an idiot, I'm going to describe what my friends and I have been talking about, and I hope to gain some insight on this.

I've come up with the thought, that;
lim x→1+ (x+1) / (x - 1) = -∞

while

lim x→1- (x+1) / (x - 1) = ∞

Now what this is telling me, is that on either side, it gets infinitesimally close to one, but it never touches one. I thought, even though if it did touch, you'd end up dividing by zero, which would net a discontinuity on the graph anyway, but in my mind, wouldn't you get a value if it did actually touch?

I'm still rather skeptical on all of it, though.

EDIT: Oh, and sorry for the formatting, new to these boards, still trying to figure everything out.
 
  • #33
HallsofIvy
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Without trying to make myself sound like an idiot, I'm going to describe what my friends and I have been talking about, and I hope to gain some insight on this.

I've come up with the thought, that;
lim x→1+ (x+1) / (x - 1) = -∞

while

lim x→1- (x+1) / (x - 1) = ∞
I presume you mean lim xx→1+ (x+1)/(x-1)= ∞

Now what this is telling me, is that on either side, it gets infinitesimally close to one, but it never touches one.
What do you mean by "it"? x? If so, then, yes, the limit refers to x being arbitrarily close to 1 but not equal to 1.

I thought, even though if it did touch, you'd end up dividing by zero, which would net a discontinuity on the graph anyway, but in my mind, wouldn't you get a value if it did actually touch?
There is a discontinuity in the graph, whether you are taking a limit or not. I don't understand your last part. Why would you think you would "get a value"?

I'm still rather skeptical on all of it, though.

EDIT: Oh, and sorry for the formatting, new to these boards, still trying to figure everything out.
 
  • #34
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Okay, I'll admit I was wrong, and that mathematically I can see where it equals one, but I'm still unsure where you would use this in real life situations. Would you be able to interchange .9~ with 1 at any given time?
 
  • #35
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Okay, I'll admit I was wrong, and that mathematically I can see where it equals one, but I'm still unsure where you would use this in real life situations. Would you be able to interchange .9~ with 1 at any given time?
Assuming that .9~ means .99999 ... with infinitely repeating 9's, yes, these are the same numbers. I don't know what this has to do with the limit you asked about, though.
 
  • #36
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You are saying that [itex]\lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n}) [/itex]
Right.
But I am saying that the limit is never actually reached, the expressions fall short by an infinitesimal amount.
The limiting value (1) is never reached for any finite value of n, that's true but immaterial. To say that 1 is the limit of 1 - 10-n means only that I can make 1 - 10-n arbitrarily close to 1 by a suitable choice of a finite value of n.

If you ask if I can get this expression within .0001 of 1, I'll tell you a value of n that does the trick. (BTW any larger value of n works even better.)

If you still aren't convinced and want me to get the expression within .0000001 of 1, I'll tell you the value of n that works.

No matter how close you want 1 - 10-n to be to 1, I can tell you the value of n that makes this true, and again all values larger than that n also work.

The idea is that you specify how close 1 - 10-n should be to 1, and I figure out the n that works. That's what is meant by saying that 1 - 10-n can be made arbitrarily close to 1.
If you treat the above variations of 1 as interchangeable then I agree that [itex]1^{\infty}[/itex] is indeterminate. However my earlier expressions give different results for these different versions of 1.
Please could someone explain this to me? What is the reason for this convention, and why do my expressions not disprove it?
Also, how can my expressions have a definite value if [itex]1^{\infty}[/itex] does not, as my expression is just a variation of [itex]1^{\infty}[/itex]?
 
  • #37
Hurkyl
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I'm still unsure where you would use this in real life situations.
One of the big, big main ideas about calculus is that there are many concepts that are difficult to describe directly, but fairly easy to describe how it is approximated by other concepts.

e.g. it is extremely difficult to give a complete and precise direct definition of the term "tangent line" -- but it is fairly easy to define it by saying it's the limit of "secant lines".
 
  • #38
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Assuming that .9~ means .99999 ... with infinitely repeating 9's, yes, these are the same numbers. I don't know what this has to do with the limit you asked about, though.

Yeah, ignore the limit. I realized that would be null for the fact of the zero in the denominator.

It seems like everything in my calculus class (high school level) has taught us that it gets infinitely close to 1, aka .9 repeating. What you're saying though, is that .9r and 1 are interchangeable, which would contradict what the teacher and textbook taught me.
How I think of it, the function 2-2x2 on the domain (-1 , 1]
It would approach x=-1 from the right infinitely close, but would never touch, correct? So the value would be .9repeating, which, you've stated is interchangeable with 1, which would give you a value at x = -1, but the restriction on the domain doesn't include -1.

I'm just trying to logic through this, I'm not attacking anyone at all.
 
  • #39
D H
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It seems like everything in my calculus class (high school level) has taught us that it gets infinitely close to 1, aka .9 repeating. What you're saying though, is that .9r and 1 are interchangeable, which would contradict what the teacher and textbook taught me.
Bah! High school math teachers.

0.999... and 1.0 are interchangeable. The value of an infinite series is not just "infinitely close" to the limit, it is equal to and indistinguishable from the limit of the series.
 
  • #40
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Bah! High school math teachers.
Especially the ones who don't have a degree in math.
 
  • #41
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It seems like everything in my calculus class (high school level) has taught us that it gets infinitely close to 1, aka .9 repeating. What you're saying though, is that .9r and 1 are interchangeable, which would contradict what the teacher and textbook taught me.
Can you give us a direct quote from your textbook? It is probably the case that we and your textbook are in 100% agreement, but you are misinterpreting what the book is saying.
I'm just trying to logic through this, I'm not attacking anyone at all.
I don't see what you are saying as attacks on us, and I don't think anyone else in this thread is, either. We're all engaged in a healthy debate with you, and are trying to set you straight on a fundamental concept of calculus.
 
  • #42
CRGreathouse
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It seems like everything in my calculus class (high school level) has taught us that it gets infinitely close to 1, aka .9 repeating. What you're saying though, is that .9r and 1 are interchangeable, which would contradict what the teacher and textbook taught me.
It's possible that your teacher is wrong. It seems less likely that your textbook is wrong. I second the request for a quote.
 
  • #43
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Can you give us a direct quote from your textbook? It is probably the case that we and your textbook are in 100% agreement, but you are misinterpreting what the book is saying.


I don't see what you are saying as attacks on us, and I don't think anyone else in this thread is, either. We're all engaged in a healthy debate with you, and are trying to set you straight on a fundamental concept of calculus.
I realize that, but I'm used to internet message boards where people get offended rather quickly and I'm just trying to avoid that, especially since I'm a new user.

My teacher, from what I understand, is that,
Limx->cf(x) = c, but when there are discontinuities within the function at x = c, that x gets infinitely (note, not arbitrarily; i know the difference between the words and I would have caught the difference) close to c, but never actually touches.

Now that I actually look at the books definition of a limit,
A formal Definition of a limit
Let f(x) be defined on an open interval about xo, excpet possibly at xo itself. We say that f(x) approaches the Limit L as x approaches xo, and write Lim x->xo = L, if for every number [tex]\epsilon[/tex] > 0, there exists a corresponding number [tex]\delta[/tex]> 0 such that for all x;
0 < |x -xo| [tex]< \delta[/tex] [tex]\Rightarrow[/tex] |f(x) - L| < [tex]\epsilon[/tex].
Okay, well I'm trying to make sense of this now, and I'm realizing we never really did any of this...we did limits, but we never dealt with delta, or finding values greater than delta...
This makes me rather nervous for college calculus, to be honest.
 
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  • #44
CRGreathouse
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My teacher, from what I understand, is that,
Limx->cf(x) = c, but when there are discontinuities within the function at x = c, that x gets infinitely (note, not arbitrarily; i know the difference between the words and I would have caught the difference) close to c, but never actually touches.
If you had said
If [tex]\lim_{x\to c}f(x)=L[/tex], f gets arbitrarily close to L in any neighborhood of c.​
then you would have been right. But:
  • "Infinitely close" is ill-defined, unlike "arbitrarily close", and
  • Even if the function has a discontinuity at x = c, there's no reason to assume it fails to "touch" in a neighborhood of c

The book's definition is the same as mine: the function gets arbitrarily close to its limit. No "infinitely", no "never actually touches".
 
  • #45
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This is honestly quite perturbing, because I realized we skipped the section with the formal limit definition in it. So I never even learned the formal definition of a limit. Could someone help me work through the definition? I'm reading over the book, but I'm still having a difficult time understanding it.

EDIT: Jesus tap dancing christ, "To show that the limit of f(x) as x -> xo actually equals L, we must be able to show that the gap between f(x) and L can be made less than any prescribed error, no matter how small by holding x close enough to xo"

Gah, I'm going to have to start reading my book instead of listening to his lecture.
 
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  • #46
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... Jesus tap dancing christ ... Gah, I'm going to have to start reading my book instead of listening to his lecture.
Many (most) people learn calculus and think they 'get it' because they can take derivatives and do the rate problems. The teachers typically slide past the hard parts because (1) they're hard to explain and (2) 0.9r of the students don't really care. You (and the folks who have been posting here) apparently are in the 1-0.9r :smile: fraction that do care and see how interesting and deep this really is. Find a copy of Berlinski's 'Tour of the Calculus' you will probably enjoy it.
 

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