On what base do we argue that $\lim_{n\to\infty}\big((\frac{2n+3}{3n+2})^n\big) = \lim_{n\to\infty}\big((\frac{2n}{3n})^n\big)$

Question

I only remember the nonstandard-legitimation:
Let $h$ be an infinite hyperreal, then:
$\lim_{n\to\infty}\big((\frac{2n+3}{3n+2})^n\big) = st\big((\frac{2h+3}{3h+2})^h\big)$
and $(\frac{2h+3}{3h+2})^h \approx (\frac{2h}{3h})^h = (\frac{2}{3})^h$, and therefore
$st\big((\frac{2h+3}{3h+2})^h\big) = st((\frac{2}{3})^h) = 0$

The extended real numbers with $\infty$ however, if I remember correctly, don't grant the above result here, as $\frac{\infty}{\infty}$ isn't defined.
On the other side, one can't pull the limes into $(\frac{2n+3}{3n+2})^n$.

How exactly does one argue here?

I'm not looking for any tricks or special ways to get the above result, just for the classic "too easy to even lead a proof" legitimization

Answer 1

Hint. Note that $$\left(\frac{2n+3}{3n+2}\right)^n=\underbrace{\left(\frac{2}{3}\right)^n}_{\to 0}\cdot \underbrace{\frac{\left(1+\frac{3}{2n}\right)^n}{\left(1+\frac{2}{3n}\right)^n}}_{\text{bounded}}.$$ You may also recall that for any real $a$, $$\lim_{n\to \infty}\left(1+\frac{a}{n}\right)^n=e^a.$$ See How to calculate $\lim \limits_{n \to \infty}(1 +\frac{a}{n})^n$?

Answer 2

We can consider that

$$\left(\frac{2n+3}{3n+2}\right)^n= e^{n\log \left(\frac{2n+3}{3n+2}\right)}\to 0$$

$$\left(\frac{2n}{3n}\right)^n= e^{n\log \left(\frac{2}{3}\right)}\to 0$$

and

$$\frac{\left(\frac{2n+3}{3n+2}\right)^n}{\left(\frac{2n}{3n}\right)^n}=\left(\frac{6n+9}{6n+4}\right)^n=\left(1+\frac{5}{6n+4}\right)^n=\left[\left(1+\frac{5}{6n+4}\right)^{\frac{6n+4}{5}}\right]^{\frac{5n}{6n+4}}\to e^{\frac56}$$