.999 does not equal 1 because

[Duncan1]

Represent .999... by (1-10^-n), limit as n>infinity.
Then
(1-10^-n)¹⁰ⁿ, limit as n>infinity = 1/e (binomial expansion)
(1+10^-n)¹⁰ⁿ, limit as n>infinity = e
(1)¹⁰ⁿ, limit as n>infinity=1

I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...
If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.


The above equation seems to converge with the number of digits of accuracy approximately equal to n.

 

[epkid08]

If you need proof that .999...=1, take a look at the infinite series that represents .999...

 

[Werg22]

If 0.99... is not equal to 1, then there's definitely a number between the two. Can you find one such number?

 

[Mute]

Represent .999... by (1-10^-n), limit as n>infinity.
Then
(1-10^-n)¹⁰ⁿ, limit as n>infinity = 1/e (binomial expansion)
(1+10^-n)¹⁰ⁿ, limit as n>infinity = e
(1)¹⁰ⁿ, limit as n>infinity=1

I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...
If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.The above equation seems to converge with the number of digits of accuracy approximately equal to n.

It doesn't work like that. You're saying

\mbox{Let}~0.\overline{9} \equiv \lim_{n \rightarrow \infty} (1 - 10^{-n}).

First, the limit of that expression is certainly 1 - no doubt about that. It's not the way one tends to define a decimal expansion but let's use it anyways. You then consider

\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}
which is NOT equal to

<br /> \lim_{n \rightarrow \infty} (0.\overline{9})^{10^n}.

The proper expression given your proposed definition for 0.\overline{9} would be

\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n} = \lim_{n \rightarrow \infty} (\lim_{m \rightarrow \infty} (1 - 10^{-m}))^{10^n},

which is a different thing altogether - there are two limits involved, instead of just the one as in your original post. This makes a very big difference. Note also that you cannot exchange the order of the limits here -that would also be an invalid operation.

 

[Count Iblis]

http://arxiv.org/abs/0811.0164

 

[Gerenuk]

The proper expression given your proposed definition for 0.\overline{9} would be
\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n} = \lim_{n \rightarrow \infty} (\lim_{m \rightarrow \infty} (1 - 10^{-m}))^{10^n},

I agree with Mute. The point is that you cannot make the single limit you had first into a double limit with a second occurence of the same infinity.

 

[rochfor1]

This has been done to death. Google is your friend.

 

[Petek]

This has been done to death. Google is your friend.

Correct. If this Wikipedia article doesn't convince you, then probably nothing ever will.

 

[Duncan1]

It doesn't work like that. You're saying

\mbox{Let}~0.\overline{9} \equiv \lim_{n \rightarrow \infty} (1 - 10^{-n}).

First, the limit of that expression is certainly 1 - no doubt about that. It's not the way one tends to define a decimal expansion but let's use it anyways. You then consider

\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}


which is NOT equal to

<br /> \lim_{n \rightarrow \infty} (0.\overline{9})^{10^n}.

The proper expression given your proposed definition for 0.\overline{9} would be

\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n} = \lim_{n \rightarrow \infty} (\lim_{m \rightarrow \infty} (1 - 10^{-m}))^{10^n},

which is a different thing altogether - there are two limits involved, instead of just the one as in your original post. This makes a very big difference. Note also that you cannot exchange the order of the limits here -that would also be an invalid operation.

However,
\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}=1/e
has a definite value other than 1. Try some values of n in a calculator.
Is this expression in error?
Does this expression not indicate that
\lim_{n \rightarrow \infty} (1 - 10^{-n}) has a value other than 1 in the above expression?
What then is the value of \lim_{n \rightarrow \infty} (1 - 10^{-n}) in the above expression?

 

[Hurkyl]

1^{+\infty} is an indeterminate form, just like 0/0, (+\infty) / (+\infty), and (+\infty) - (+\infty).


If you can prove 0.999... is unequal to 1, then you have found an inconsistency in the arithmetic of integers.

 

[Redbelly98]

To the OP:

What is the decimal representation of 1/3?
Do you accept that 1/3 = 0.333...?

If so, multiply x3 on both sides of that equation and see what you get.

 

[Count Iblis]

To the OP:

What is the decimal representation of 1/3?
Do you accept that 1/3 = 0.333...?

If so, multiply x3 on both sides of that equation and see what you get.

If 0.999999... is questioned to be equal to 1, then why would one accept that 0.33333333... equals 1/3?


I think the answer to ".999... does not equal 1 because..." should be that the number system has not been specified, as pointed out here:

http://arxiv.org/abs/0811.0164

So long as the number system has not been specified, the students' hunch that .999... can fall infinitesimally short of 1, can be justified in a mathematically rigorous fashion.

 

[Redbelly98]

If 0.999999... is questioned to be equal to 1, then why would one accept that 0.33333333... equals 1/3?

That occurred to me. Hence my initial question, "What is the decimal representation of 1/3?" Most people are okay with saying 0.333... in answer to that, and don't question it.
However, the perception that 0.999... looks differently than 1.000... is what distinguishes the two examples in many people's minds.

 

[Redbelly98]

So long as the number system has not been specified, the students' hunch that .999... can fall infinitesimally short of 1, can be justified in a mathematically rigorous fashion.

I don't understand -- do we not have an agreed-upon number system?

 

[Hurkyl]

I don't understand -- do we not have an agreed-upon number system?

The author of that PDF is writing it pretty much specifically to justify the 0.999... != 1 hypothesis. The main loophole he exploits is that we haven't given a fully precise definition of the number system we are teaching them -- and so he's filling in the missing details in a... peculiar... fashion.

That is not to say there is no merit in non-standard analysis -- but it's like the author is specifically trying to prevent students from understanding what an infinite decimal means.

 

[Tac-Tics]

I don't understand -- do we not have an agreed-upon number system?

Most people have a terrible understanding of the real numbers. It isn't usually until college (more likely, never) that people learn about the reals as the completion of the rational number system. The vast majority of people thing an irrational number's defining characteristic is that the decimal expansion doesn't repeat. And they believe that you can add two irrational numbers together in the same way you add rationals, using the algorithm they learned in elementary school -- despite the fact that it obviously doesn't terminate!

The peculiar thing isn't that people don't understand numbers, it's that they so fiercely stick to their broken notions of numbers. They won't even entertain the idea that what they learned was incomplete. After all, business and science are both done with rational numbers! And there's a strange belief (in both math and science) that there's only one correct way these kinds of things can work. But that's the beauty of math! You can make up the rules to be whatever you want as long as you strictly adhere to them.

 

[lurflurf]

Most people have a terrible understanding of the real numbers. It isn't usually until college (more likely, never) that people learn about the reals as the completion of the rational number system. The vast majority of people thing an irrational number's defining characteristic is that the decimal expansion doesn't repeat. And they believe that you can add two irrational numbers together in the same way you add rationals, using the algorithm they learned in elementary school -- despite the fact that it obviously doesn't terminate!

The peculiar thing isn't that people don't understand numbers, it's that they so fiercely stick to their broken notions of numbers. They won't even entertain the idea that what they learned was incomplete. After all, business and science are both done with rational numbers! And there's a strange belief (in both math and science) that there's only one correct way these kinds of things can work. But that's the beauty of math! You can make up the rules to be whatever you want as long as you strictly adhere to them.

Most people do not need a good understanding of real numbers. Real numbers are quite obviously a completion of the rationals, they are the numbers between rational like sqrt(2). It is perfectly valid to take nonterminating decimal expansion as a defining characteristic of irrational numbers. Irrationals are aded the same way as rationals, or at least an infinite process analog. I think the trouble is these people (though most of them know .(9)=1 and just want to start stupid threads) do not realize a infinite decimal expansion is inherently a real number. Arguing .(9) is not real is like arguing 17/19 is not rational. One could (and these crack pots do) devise a number system different than the reals (and having horrible properties) but represented by infinite decimal expansions in which two representations that represent the same real number represent different crackpot numbers. The two sides of this "debate" are not even talking about the same thing. We have
Crackpot: In the crackpot number system clearly .(9)!=1.(0)
!Crackpot: In the real number system clearly .(9)=1.(0)
They are both right.

 

[Mute]

However,
\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}=1/e
has a definite value other than 1. Try some values of n in a calculator.
Is this expression in error?

No, that expression is perfectly correct.

Does this expression not indicate that
\lim_{n \rightarrow \infty} (1 - 10^{-n}) has a value other than 1 in the above expression?

No, it absolutely does not indicate that. \lim_{n \rightarrow \infty} (1 - 10^{-n}) = 1. Your error is the following: consider the expression

(1-10^{-n})^{10^m}.

If I take the limit n \rightarrow \infty this becomes (1)^{10^m} = 1 for any finite value of m. If I instead keep n fixed at some finite value and take m \rightarrow \infty then because 1-10^{-n} &lt; 1, the expression tends to zero. If I set m = n and then take the limit n \rightarrow \infty I get the value of 1/e. You're mixing up these limiting processes which gives you different results depending on how you take the limit.

 

[Hurkyl]

a infinite decimal expansion is inherently a real number.

I'm possibly nitpicking a bit, but:
. an infinite decimal expansion is a real number only because we have chosen to use decimal notation to express real numbers
. we only use the left-finite expansions

 

[Count Iblis]

Most real numbers are not formally describable (the set of formally describable numbers must obviously be countable), so you can question if real numbers really exist.

 

[Hurkyl]

Most real numbers are not formally describable (the set of formally describable numbers must obviously be countable), so you can question if real numbers really exist.

A common assertion, but with subtle precision issues!

If "really exist" is to mean the mathematical existence predicate, then your hypothesis here is irrelevant.

However, if you are saying something about the real world, then the implicit conclusion you made no longer follows.

e.g. Skolem's paradox demonstrates that, while internally to a (first-order) model of set theory the real numbers are not countable, that need not remain true in an external context. In particular, there exists a countable model of (first-order) set theory.

 

[Werg22]

Most real numbers are not formally describable (the set of formally describable numbers must obviously be countable), so you can question if real numbers really exist.

That's really akin to asking whether the geometrical line, infinite sets, ordered pairs, functions or relations really exist. Do ideas exist?

 

[Mark44]

It is perfectly valid to take nonterminating decimal expansion as a defining characteristic of irrational numbers.

Did you really mean to say that? If so, then .636363636363... and .222222... are irrational, despite the fact that they can be represented respectively as 7/11 and 2/9, which are clearly rational numbers.

 

[Duncan1]

No, that expression is perfectly correct.



No, it absolutely does not indicate that. \lim_{n \rightarrow \infty} (1 - 10^{-n}) = 1. Your error is the following: consider the expression

(1-10^{-n})^{10^m}.

If I take the limit n \rightarrow \infty this becomes (1)^{10^m} = 1 for any finite value of m. If I instead keep n fixed at some finite value and take m \rightarrow \infty then because 1-10^{-n} &lt; 1, the expression tends to zero. If I set m = n and then take the limit n \rightarrow \infty I get the value of 1/e. You're mixing up these limiting processes which gives you different results depending on how you take the limit.

Thank you for trying to explain this to me.
Clearly I am looking at the case where m=n for limits. What is the rule for separating limits? For example xsin(1/x) limit x->infinity or sin(y)/y limit y->0 would be undefined if you separate the limits. Or indeed my above examples would be undefined if you separate the limits.

by taking m=n, I am trying to make an example where \lim_{n \rightarrow \infty} (1 - 10^{-n}) != 1. (albiet by an infinitesimal amount.) Are you saying this is not allowed by some rule or convention? Please could you explain that rule?
Duncan Irvine.

 

[Duncan1]

1^{+\infty} is an indeterminate form, just like 0/0, (+\infty) / (+\infty), and (+\infty) - (+\infty).


If you can prove 0.999... is unequal to 1, then you have found an inconsistency in the arithmetic of integers.

I disagree with 1^{+\infty} is indeterminate.
If my initial equations are correct then 1^{+\infty} is bounded by e, 1/e.
If the equations are changed to
\lim_{n \rightarrow \infty} (1 + x(10^{-n}))^{10^n}=e^x

Then for x=+1 and -1, 1^{+\infty} has a value between e^x and e^-x.
limit as x->0 gives a value between e^+0 and e^-0=e^0=1.


An interesting variant is:
\lim_{n \rightarrow \infty} (1 + i(pi/2)(10^{-n}))^{10^n}=i

An infinitesimal imaginary component can also be expanded into a complex number.

Duncan Irvine.

 

[Redbelly98]

I disagree with 1^{+\infty} is indeterminate.
If my initial equations are correct then 1^{+\infty} is bounded by e, 1/e.

Here's a counterexample to that statement:

lim_n→∞ [(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.

 

[Count Iblis]

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).

 

[Duncan1]

Here's a counterexample to that statement:

lim_n→∞ [(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

lim_n→∞ [(1+10^-n)^2·10n] = e²​

The first term is 1^{+\infty} 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, \(1 + 10^{-n} >1. Your equation uses \(1 + 10^{-n} in place of 1, and demonstrates that \(1 + 10^{-n} >1 for infinite n at least for this expression.

 

[Mark44]

I disagree, if you do a binomial expansion of

lim_n→∞ [(1+10^-n)^2·10n] = e²​

The first term is 1^{+\infty} 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 1^{\infty} has a definite value. To the contrary, he is demonstrating an expression of this indeterminate form that is a counterexample to your assertion that 1^{\infty} is bounded by 1/e and e.

All of the indeterminate forms, of which [1^{\infty}] 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], [\infty/\infty], and [\infty - \infty].

 

[Duncan1]

Redbelly98 is not saying that 1^{\infty} has a definite value. To the contrary, he is demonstrating an expression of this indeterminate form that is a counterexample to your assertion that 1^{\infty} is bounded by 1/e and e.

All of the indeterminate forms, of which [1^{\infty}] 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], [\infty/\infty], and [\infty - \infty].

You are saying that \lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n})
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 1^{\infty} 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 1^{\infty} does not, as my expression is just a variation of 1^{\infty}?

 

[Hurkyl]

You are saying that \lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n})
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​

 

[cheezeitz]

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.

 

[HallsofIvy]

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.

 

[cheezeitz]

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?

 

[Mark44]

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.

 

[Mark44]

You are saying that \lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n})

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 1^{\infty} 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 1^{\infty} does not, as my expression is just a variation of 1^{\infty}?

 

[Hurkyl]

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".

 

[cheezeitz]

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.

 

[D H]

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.

 

[Mark44]

Bah! High school math teachers.

Especially the ones who don't have a degree in math.

 

[Mark44]

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.

 

[CRGreathouse]

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.

 

[cheezeitz]

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,
Lim_x->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 x_o, excpet possibly at x_o itself. We say that f(x) approaches the Limit L as x approaches x_o, and write Lim _x->xo = L, if for every number \epsilon > 0, there exists a corresponding number \delta> 0 such that for all x;
0 < |x -x_o| &lt; \delta \Rightarrow |f(x) - L| < \epsilon.

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.

 

[CRGreathouse]

My teacher, from what I understand, is that,
Lim_x->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 \lim_{x\to c}f(x)=L, 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".

 

[cheezeitz]

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 -> x_o 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 x_o"

Gah, I'm going to have to start reading my book instead of listening to his lecture.

 

[gmax137]

... 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 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.