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

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?

• The inequality $x - 1 < \lfloor{x}\rfloor < x$ is not true. You should be looking at $x - 1 < \lfloor{x}\rfloor \leq x$ instead. – Clement Yung Dec 28 '19 at 14:15
• $3e^x-1<[3e^x]\leq 3e^x$ similarly $2e^x-1<[2e^x]\leq 2e^x$ now can you squeeze $f(x)$ between two values whose limit is $2/3$? – kingW3 Dec 28 '19 at 14:18

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.

• 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! – Invisible Dec 28 '19 at 14:26

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.

• Just because $\lim a=\lim b=\infty$ and $\lim c=\lim d=\infty$ doesn't imply $\lim a/c=\lim b/d$ – kingW3 Dec 28 '19 at 14:22
• 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 – Rhys Hughes Dec 28 '19 at 14:28
• 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 $2$. – kingW3 Dec 28 '19 at 15:24
• "drowns out the constant term" is sufficient in my opinion – Rhys Hughes Dec 28 '19 at 16:10