e as limit with a rational sequence instead of n Hello and happy New Year to the Stackexchange community. I had trouble solving this seemingly trivial problem and it would be great if you could help me out.
$$\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$$
This definition of e is given and I want to show that the following identitiy is true as well.
$$\lim_{n \to \infty} (1+\frac{1}{q_{n}})^{q_{n}} = e$$
with $q_{n}$ as a positive rational sequence which diverges to $+\infty$ (therefore, $\lim_{n \to \infty} q_{n}=+\infty$).

I figured that since
$$\lim_{n \to \infty} n = +\infty = \lim_{n \to \infty} q_{n}$$
I could move $\lim_{n \to \infty}$ inside the given expression, so that we have
$$\lim_{n \to \infty}(1+\frac{1}{q_{n}})^{q_{n}}=(1+\frac{1}{\lim_{n \to \infty} q_{n}})^{\lim_{n \to \infty}q_{n}} = (1+\frac{1}{\lim_{n \to \infty} n})^{\lim_{n \to \infty} n} = \lim_{n \to \infty}(1+\frac{1}{n})^{n}=e.$$
However, apperantly, I need to show that the sequence is continuous in order to do this and my course didn't introduce continuity yet. Of course, I could first prove the therom needed to move the limit inside the expression and then show the sequence is continous, but I feel like there is a much more easier and elegant solution to this.
I'm sorry if this is a duplicate of a previously answered question (at least I couldn't find one) and any insight would be appreciated. Thank you very much.

Update — I think I managed to solve it by brute force, but it took me hours and my proof is 2+ pages long. I'm sure that my proof was not the intended one and there must be a simpler one. Anyway, thank you for the help and when I have time, I will update this question with my solution as reference.
 A: It is not much difficult to establish what you seek. The function $$f(x) = \left(1 + \frac{1}{x}\right)^{x}$$ for $x \in \mathbb{Q}^{+}$ is an increasing function. This is easy to prove via Bernoulli's inequality. Let $x, y$ be positive rationals with $x > y$ then we have $$\left(1 + \dfrac{1}{x}\right)^{x/y} \geq 1 + \frac{x}{y}\cdot\frac{1}{x} = 1 + \frac{1}{y}$$ so that $f(x) \geq f(y)$.
Let $q_{n}$ be a sequence of rationals tending to $\infty$ and let $s_{n} = \lfloor q_{n}\rfloor$ and $t_{n} = s_{n} + 1$. Clearly we can see that from a certain value of $n$ the sequences $s_{n}, t_{n}$ are made of positive integers only and both tend to $\infty$. Now we know that $f(n) \to e$ as $n \to \infty$ and hence $f(s_{n}), f(t_{n})$ both tend to $e$ as $n \to \infty$. Since $s_{n}\leq q_{n}\leq t_{n}$ it follows that $f(s_{n}) \leq f(q_{n})\leq f(t_{n})$ and by Squeeze theorem $f(q_{n}) \to e$ as $n \to \infty$.

Note: The above proof requires the version of Bernoulli's inequality with rational exponents.
A: (long comment)
I suggest a way(not the only way) here. First you can treat $\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$ as a limit of a function $f:\Bbb R\to\Bbb R$, rather than a sequence, to infinity. So you may write $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$ instead.
Then, if you have admit that $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$ to be true, then you can directly use the famous theorem, Sequential Characterization of Limits(in particular, the version that related to limits as $x$ goes to $\infty$, if you don't know, just google it!). Then you can instantly get that $\lim_{n \to \infty} (1+\frac{1}{q_{n}})^{q_{n}} = e$ for all rational sequence $q_n$ that diverges to $\infty$.
PS: I'm afraid that if you are only allowable to admit $\lim_{n \to \infty}(1+\frac{1}{n})^{n}=e$ but not $\lim_{x \to \infty}(1+\frac{1}{x})^{x}=e$, then your claim can not be proved, though I'm not sure about this.
