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Okay, so I read this somewhere that,

$$ \lim_{x \to 0^+} \left[ \frac{\sin x}{x} \right] = 0 $$ Where, [] denotes the greatest integer function.

But, on the other hand, this is also true,

$$ [0.9999...] = 1 $$

Aren't these two contradictory? I mean if,

$$ \lim_{x \to 0^+} \frac{\sin x}{x} = 1 $$

and $$ 0.9999... = 1 $$

Then why is the greatest Integer function behaving differently for these two functions?

Asaf Karagila
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AnonMouse
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3 Answers3

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$0.999\ldots$ isn't a sequence / function you take the limit of. It is a fixed number, and it's equal to $1$ in value (more strictly: If you allow it to represent a value, then any value except $1$ will get you into inconsistencies). Round it down all you like, that doesn't change. On the other hand, we do have $$ \lim_{n \to \infty}\left[0.\underbrace{999\ldots99}_{n\text{ times}}\right] = 0 $$ As for why $\lim_{x \to 0^+} \left[ \frac{\sin x}{x} \right] = 0$, that's simply because for all non-zero $x$, we have $\left[ \frac{\sin x}{x} \right] = 0$, so of course the limit is going to be $0$ as $x \to 0$.

Asaf Karagila
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Arthur
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  • Thanks a lot for your answer. But I am a bit confused, is it wrong to write that $ 0.999... < 1 $? – AnonMouse Jul 22 '16 at 11:42
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    @BrahmnoorSingh Yes, $0.999\ldots < 1$ is wrong. The correct statement is $0.999\ldots = 1$. – Arthur Jul 22 '16 at 11:43
  • @BrahmnoorSingh Have a look at :http://math.stackexchange.com/q/11/321264 – StubbornAtom Jul 22 '16 at 11:44
  • @BrahmnoorSingh: Especially see this answer. –  Jul 22 '16 at 11:46
  • Thanks a lot. I did read these answers, and it might take me some time to wrap my head around those concepts, but there is one thought that's constantly bothering me. Is it not a limiting concept that we are doing here? As @Arthur wrote in the answer, $ \lim_{n \to \infty}\left[0.\underbrace{999\ldots99}{n\text{ times}}\right] = 0 $. But why is it that 0.999... and $ \lim{n \to \infty}\left[0.\underbrace{999\ldots99}_{n\text{ times}}\right] $ are different? – AnonMouse Jul 22 '16 at 11:51
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    @BrahmnoorSingh There is a big difference between the two expressions $$\left[\lim_{n \to \infty}0.\underbrace{999\ldots99}{n\text{ times}}\right] \quad\text{and}\quad \lim{n \to \infty}\left[0.\underbrace{999\ldots99}_{n\text{ times}}\right] $$ The first one is what you refer to as $[0.999\ldots]$, while the second one is a sequence of numbers, all equal to $0$, so their limit is clearly $0$. – Arthur Jul 22 '16 at 11:52
  • @BrahmnoorSingh Do you see the difference between "$\lim_{x \to 1^-}[x]$" and "$[1]$"? Because I think, at it's core, that's what you're asking about. – Arthur Jul 22 '16 at 11:56
  • @Arthur The picture is becoming clearer, and I think I understand it better. I think at it's core, I am not able to comprehend that $ 0.999... = 1 $. Because once that is clear, than everything else is quite explanatory. – AnonMouse Jul 22 '16 at 12:00
  • @BrahmnoorSingh , what will be [0.99999...] then ? – Jdeep Aug 14 '20 at 16:12
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There are too many red herrings in your question. Let's remove them entirely.

Consider the function $F$ such that $F(x)=0$ for all $x<1$ and $F(1)=1$. This function is not continuous at $1$. More specifically, $$\lim_{x\to 1^{-}} F(x)=0$$ because it is the limit of a constant function which always returns $0$. This is despite the obvious fact that $\lim_{x\to 1^{-}}x=1$.

This is the same situation in your question. You compose the $\frac{\sin x}x$ function with a non-continuous function which ensures that the value becomes a constant $0$; and then you compare it to a different limit.

Asaf Karagila
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$$ 0.9999... = 1 $$ is perfectly true, but when you take the limit, you never evaluate the function at exactly $0$, as it is undefined, so you never have infinitely many $9$s, and the above equality does not enter into play.