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

Question

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?

Answer 1

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.

Answer 2

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.