$\lim_{x\to0}x=0$
$\lim_{x\to1}x=1$
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?
${0.9999\dots}$ 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.