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This is for all $x\in\mathbb{R}$ and $x\notin\mathbb{Z}$ because it equals $1$ for all positive integers.

I was just messing around with floor and ceiling functions in Desmos when I came upon this. I have not been able to prove this and it is at the moment only an observation. To me, this makes absolutely no sense and I have no clue how to begin solving this. When typing it into WolframAlpha, it just gave me an answer of $1$ because it did not exclude integers. It also gave me a partial product formula of $\left(\frac{\lceil x\rceil}{\lfloor x\rfloor}\right)^n$ but this seems like it would not work for $n=\infty$. This is what it looks like in Desmos: enter image description here Is anyone able to explain this to me? How do you even prove this? Thanks for the help in advance!

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    $\begingroup$ @aschepler The OP's title doesn't match their question $\endgroup$
    – FShrike
    Jun 9, 2023 at 18:04
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    $\begingroup$ OP's attached picture clearly says what the author means, just the title is not well written. @MartinR $\endgroup$
    – MathFail
    Jun 9, 2023 at 18:09
  • $\begingroup$ Thanks for fixing the question. I am not quite sure why mine was incorrect, but I appreciate it nonetheless. $\endgroup$ Jun 9, 2023 at 18:10
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    $\begingroup$ It's clear to me what OP means and what MathFail said. OP just has confused their Desmos graph with their title $\endgroup$
    – FShrike
    Jun 9, 2023 at 18:10
  • $\begingroup$ @FShrike Thanks. Would you mind explaining what was different between the original question and what I meant. I thought because I used the same equation that I used in the Desmos, it would be the same. Sorry if this doesn't make sense, I am just quite confused. $\endgroup$ Jun 9, 2023 at 18:13

1 Answer 1

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Let $k=\lfloor x\rfloor$

$$\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\prod\limits^{k}_{n=1}\frac{k+1}{k}=\left( 1+\frac1k\right)^k$$

therefore,

$$\lim_{x\to\infty\land x\notin\mathbb N}\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\lim_{k\to\infty}\left( 1+\frac1k\right)^k=e$$

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    $\begingroup$ What is $n$? This makes as little sense to me as the OP did. $\endgroup$ Jun 9, 2023 at 18:04
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    $\begingroup$ @TedShifrin the multiplicand just doesn't depend on $n$ so we get a product of constants $\endgroup$
    – FShrike
    Jun 9, 2023 at 18:05
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    $\begingroup$ Note OP is not dependent on $n$, that's why it gets $e$ $\endgroup$
    – MathFail
    Jun 9, 2023 at 18:05
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    $\begingroup$ Note that when $x$ is an integer, this evaluates to $1$, not $e$, so the limit isn’t actually defined. $\endgroup$
    – Eric
    Jun 9, 2023 at 21:17
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    $\begingroup$ Read the first sentence of OP, the author said that. @Eric $\endgroup$
    – MathFail
    Jun 9, 2023 at 22:18

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