Checking my proof that $\lim_{n \to \infty} 1/(2n - 3) = 0$ My goal is to prove that the sequence $(x_n)$ defined as $x_n := \frac{1}{2n-3}$ converges and the limit is $0$. In addition to determining if my proof is correct I also wanted to know if there is an alternative way to do the proof, maybe with an inequality that holds for all $n \in \mathbb{N}$, still just using the definition of the limit of a sequence.
My proof:
Let $\varepsilon > 0$. By the Archimedean Property there exists some $K_{*} \in \mathbb{N}$ such that $1/\varepsilon < K_{*}$, therefore $1/K_{*} < \varepsilon$. Let $K = \max\{K_{*}, 3\}$ and suppose $n \geq K$. Then
$$\bigg|\frac{1}{2n-3} - 0 \bigg| = \bigg|\frac{1}{2n-3}\bigg| \leq \bigg|\frac{1}{n}\bigg| = \frac{1}{n} \leq \frac{1}{K} < \varepsilon,$$
where the first inequality holds for all $n \geq 3$.
 A: Proof makes sense to me. Another way you can prove it is that if you already proved $\frac{1}{n}\rightarrow 0$, note your sequence is a subsequence of $\frac{1}{n}$ (excluding the first element), and thus it converges to zero since all subsequences of a convergent sequence converge to the same thing.
A: Good proof! But here's a similar way to do it, using the theorem that if $ |s_n - s| \le a_n $, where $a_n \to 0$, then $s_n \to s$. (I'll refer to it as Theorem 3.1.9 in the proof because that's what it was named when I first learned it.)
Claim: $\lim(x_n) = 0$.
Proof. We have: $|\frac{1}{2n-3}-0| = \frac{1}{2n-3} \lt \frac{1}{n} \to 0$ for $n \gt 3$. Therefore, by Theorem 3.1.9, $\lim(\frac{1}{2n-3}) = 0$. $$\tag*{$\blacksquare$}$$
A: Since $\lim\limits_{x \to 0^+} \frac{1}{x} = +\infty$, we can have $u = 1/n$ and proving $$\lim_{u \to 0^+} \frac{3u}{2u + 3} = 0$$ which is the same problem. One way of proving this is to have $\delta \leq 1$, and we'll have $\delta = \min\left\{1, \frac{\varepsilon}{3}\right\}$.
