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Prove that:

$\displaystyle\lim_{x\to \infty}\frac{\lfloor 3e^x\rfloor+2}{\lfloor 2e^x\rfloor+1}=\frac{3}{2}$

My attempt: Let $f:\mathbb R\to S\subset\mathbb Q$ $$f(x)=\frac{\lfloor 3e^x\rfloor+2}{\lfloor 2e^x\rfloor+1}.$$ I was a bit confused because I thought I was required to prove the statement by choosing an appropriate $\epsilon\;\&\;\delta$ and plugging the limit into: $$x\in\langle c-\epsilon,c+\epsilon\rangle\implies|f(c)-L|<\delta\;$$ which went unsuccessfully since $f$ is discontinuous.

I considered rewriting the numerator & the denominator and use the fact: $$x-1<\lfloor x\rfloor\leq x,$$ but I was wondering if I could simply remove the, in this case, 'insignificant' constants $1\;\&\;2.$ However, $\displaystyle\lim_{x\to\infty}\frac{\lfloor3e^x\rfloor}{\lfloor2e^x\rfloor}$ doesn't seem completely simplified. Should I apply the Stolz-Cesaro theorem?

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    $\begingroup$ The inequality $x - 1 < \lfloor{x}\rfloor < x$ is not true. You should be looking at $x - 1 < \lfloor{x}\rfloor \leq x$ instead. $\endgroup$
    – Clement Yung
    Dec 28, 2019 at 14:15
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    $\begingroup$ $3e^x-1<[3e^x]\leq 3e^x$ similarly $2e^x-1<[2e^x]\leq 2e^x$ now can you squeeze $f(x) $ between two values whose limit is $2/3$? $\endgroup$
    – kingW3
    Dec 28, 2019 at 14:18

2 Answers 2

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Hint: Since: $$ x - 1 \leq \lfloor{x}\rfloor \leq x $$ We have: $$ \frac{3e^x + 1}{2e^x + 1} \leq \frac{\lfloor{3e^x}\rfloor + 2}{\lfloor{2e^x}\rfloor + 1} \leq \frac{3e^x + 2}{2e^x} $$ Now apply Squeeze Theorem.

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  • $\begingroup$ thank you very much! I was really clumsy with this, but it looks rather beautiful, especially when I compare the first and the third term. Thank you once again! $\endgroup$
    – Invisible
    Dec 28, 2019 at 14:26
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Both $\lfloor{3e^x}\rfloor$ and $\lfloor{2e^x}\rfloor \to \infty$, so:

$$\lim_{x\to\infty}\lfloor{3e^x}\rfloor+2=\lim_{x\to\infty}\lfloor{3e^x}\rfloor$$ $$\lim_{x\to\infty}\lfloor{2e^x}\rfloor+1=\lim_{x\to\infty}\lfloor{2e^x}\rfloor$$

Then we get: $$\lim_{x\to\infty}\frac{\lfloor{3e^x}\rfloor+2}{\lfloor{2e^x}\rfloor+1}=\lim_{x\to\infty}\frac{\lfloor{3e^x}\rfloor}{\lfloor{2e^x}\rfloor}=\lim_{x\to\infty}\frac32=\frac32$$

It works in the same way as if it were $\frac{3x^2+2}{2x^2+1}$, the increasing $x^2$ (or $e^x$) drowns out the constant term.

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    $\begingroup$ Just because $\lim a=\lim b=\infty$ and $\lim c=\lim d=\infty$ doesn't imply $\lim a/c=\lim b/d$ $\endgroup$
    – kingW3
    Dec 28, 2019 at 14:22
  • $\begingroup$ True, however the relative distance between $a$ and $b$ (and also between $c$ and $d$) constricts to almost nothing as $x\to\infty$. That's enough here $\endgroup$
    – Rhys Hughes
    Dec 28, 2019 at 14:28
  • $\begingroup$ Yeah, though I think you should be explicit about that, a good way would be for example to divide by $e^x$ top and bottom and then it's easy to prove that the top goes to $3$ and the bottom goes to $2$. $\endgroup$
    – kingW3
    Dec 28, 2019 at 15:24
  • $\begingroup$ "drowns out the constant term" is sufficient in my opinion $\endgroup$
    – Rhys Hughes
    Dec 28, 2019 at 16:10

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