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$\lim_{x\to0}x=0$

$\lim_{x\to1}x=1$

Limit is unreachable, isn't that mean for x→0 can never be 0?

That's why infinitesimal isn't 0, because its limit is 0.

(By Wikipedia:

infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zero.)

But if apply the same cases,

x→1 has a limit of 1, and $x$ can never be 1.

So what's the difference between the $x$ here and 0.999... ?

Both of them are approaching 1, not equal to 1 and greater than all the number smaller than 1.

If this is true then 1 should only be the limit of 0.999... but not equal.

Just as infinitesimal doesn't exist on standard real number system, shouldn't 0.999... also not exist in standard real number system but rather than equal to 1?

Where am I wrong?

Really looking forward to replies.

Can't really accept 0.999...=1 from such reason:

there is no real number between 0.999... and 1, so they are equal.

Isn't this the idea and concept of limit?

David C. Ullrich
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danny
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    In fact $1$ is the limit of the sequence $0.9,0.99,0.999,0.9999,\cdots$ – Peter Apr 04 '21 at 18:03
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    Does this answer your question? Does $[0.9999....]=1$?. This question is even a better duplicate. – Dietrich Burde Apr 04 '21 at 18:04
  • "That's why infinitesimal isn't 0, because its limit is 0." No. An infinitesimal is just one (non-standard) number, not a sequence, so there's simply no such thing as "its limit". – David C. Ullrich Apr 04 '21 at 18:35
  • So the $x$ in $lim(x→0)x=0$ is not as same as the infinitesimal? But if it is a non-standard number, then how come $0.999...$ would be a fixed value. Because both of them are approaching to a number so I thought they should both be same that not equal to a finite number ? – danny Apr 04 '21 at 18:57

1 Answers1

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${0.9999\dots}$ by definition, is a limit. It's the limit of the sequence $$ 0.9,0.99,0.999,\dots $$ this limit is $1$. And so you don't need to specify "the limit of ${0.999\dots}$" since when we talk about ${0.999\dots}$ we are specifically referring to the number which is the limit of the sequence ${0.9,0.99,0.999,\dots}$ which is $1$. So ${0.999\dots=1}$.

EDIT: I decided to make an edit to address $2$ things.

Firstly, OP (@danny) has asked for the proof that the limit of the sequence is $1$ and ${0.999\dots}$. Firstly, the limit is ${0.999\dots}$ by definition (we have defined it to be this). Now - let's prove the limit is $1$ also.

Label the sequence as ${x_n}$: $$ x_n = 0.\underbrace{999\dots 9}_{n\text{ nines}} $$ we wish to find ${\lim_{n\rightarrow\infty}x_n}$. Well, simply notice that $$ 1-x_n = 0.\underbrace{000\dots 0}_{n-1\text{ zeroes}}1 = \frac{1}{10^n} $$ you can notice that $$ \lim_{n\rightarrow\infty}(1-x_n) = \lim_{n\rightarrow\infty}\frac{1}{10^n} = 0 $$ this directly implies $$ \lim_{n\rightarrow\infty}x_n = 1 $$ as required.

The other thing I'd like to address is comments left by @Tyma Gaidash and @David C. Ullrich. @Tyma Gaidash referenced the famous derivation that ${0.999\dots=1}$: $$ \begin{array}{ccc} &n:=0.999\dots&\\ \Rightarrow&10n = 9.999\dots&\\ \Rightarrow&10n - n = 9&\\ \Rightarrow&n=1&\\ \Rightarrow&1=0.999\dots& \end{array} $$ to which @David C. Ullrich. debunked as a valid proof. It's true that this is not really a "proof", since just as @David C. Ullrich. said - this would assume that the notation ${n=0.999\dots}$ makes sense as an algebraic object in the first place, and that the manipulations above also make sense and hold. We have to first define what the notation ${0.999\dots}$ means. You could take a definition and show the steps above to be valid, but it would be overcomplicated. Although it's definitely a nice way to informally look at it.

Riemann'sPointyNose
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  • Thanks for your answer and sorry for my mistakes. If the limit of the sequence is $0.999...=1$, then the sequence shouldn't be the limit itself? Is that mean the limit of the sequence is $0.999...$ but also $1$ so we said 0.999... equal to 1? – danny Apr 04 '21 at 18:11
  • @danny what I am saying is that ${0.999\dots}$ is defined as the limit of ${0.9,0.99,\dots}$. We can prove the limit of this sequence is $1$. So ${0.999\dots=1}$ since they are both the limit of this sequence (and limits are unique). Do you see what I mean? I am not defining ${0.999\dots}$ as the sequence itself, I am defining it literally as the limit of the sequence – Riemann'sPointyNose Apr 04 '21 at 18:16
  • @danny also don't be sorry, studying Mathematics is impossible to do without mistakes! – Riemann'sPointyNose Apr 04 '21 at 18:19
  • Oh I get what you mean! Thanks a lot! – danny Apr 04 '21 at 18:21
  • Hello also there is the traditional proof:

    Here is a proof that .999…=1:

    Let 10n=9.999... then, n=.999...

    This means 10n-n=9.999-.999=9n=9.

    Dividing by nine gets us that n=$\frac 99$=1

     – Тyma Gaidash Apr 04 '21 at 18:22
  • I head this before but I just think so weird - I thought it should act like infinitesimal that not equal to any real number instead of equal to 1. Cause obviously they are totally different from looking: a finite number and a number with infinite 9. – danny Apr 04 '21 at 18:30
  • @TymaGaidash That "traditional proof" as close to no proof at all; you're assuming that the notation 0.999... actually makes sense – David C. Ullrich Apr 04 '21 at 18:37
  • @Riemann'sPointyNose Sorry but would I ask for where to find the proof of limit of the series be both $1$ and $0.999...$? – danny Apr 04 '21 at 19:00
  • @danny yes that's no problem, I can quickly add it too my answer – Riemann'sPointyNose Apr 04 '21 at 19:10
  • @DavidC.Ullrich thanks for the clarification. This notation was used in the question defined as a limit. Can it not be in my answer also defined as a limit from the original question or am I wrong? – Тyma Gaidash Apr 04 '21 at 19:15
  • @DavidC.Ullrich and Tyma Gaidash, I have added a little about your comments in my answer – Riemann'sPointyNose Apr 04 '21 at 19:22