1
$\begingroup$

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?

$\endgroup$
0

3 Answers 3

3
$\begingroup$

$0.999\ldots$ 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 $x \to 0$.

$\endgroup$
9
  • $\begingroup$ Thanks a lot for your answer. But I am a bit confused, is it wrong to write that $ 0.999... < 1 $? $\endgroup$
    – AnonMouse
    Jul 22, 2016 at 11:42
  • 1
    $\begingroup$ @BrahmnoorSingh Yes, $0.999\ldots < 1$ is wrong. The correct statement is $0.999\ldots = 1$. $\endgroup$
    – Arthur
    Jul 22, 2016 at 11:43
  • $\begingroup$ @BrahmnoorSingh Have a look at :math.stackexchange.com/q/11/321264 $\endgroup$ Jul 22, 2016 at 11:44
  • $\begingroup$ @BrahmnoorSingh: Especially see this answer. $\endgroup$
    – user170039
    Jul 22, 2016 at 11:46
  • 2
    $\begingroup$ @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 $0$. $\endgroup$
    – Arthur
    Jul 22, 2016 at 11:52
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$$ 0.9999... = 1 $$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.