In my lecture notes there is a side note to the proof for the example below that the last inequality $\frac{14}{n}\lt{\epsilon}$ in the equation is not always true and only holds under the condition that $n\gt \frac{14}{\epsilon}$, hence the sentence continues with the condition "provided $n\gt \frac{14}{\epsilon}$".
I'm confused and can't see this point. Can someone please explain to me when this does not hold, so that we proceed with the following sentence. After finding a suitable inequality for $n_0$ in other examples I usually just proceed with "take $n_0 \in \mathbb{N}$, with $n_0\gt ...$ ".
Example:
Let $(x_n)_{n=1}^{\infty}$ be given by $$x_n=\frac{3n-2}{n+4}.$$ show that $x_n\rightarrow{3}$ as $n\rightarrow{\infty}$, directly from the definition.
Solution:
Let $\epsilon \gt{0}$. For $n\in \mathbb{N}$, we have $$\left\lvert {\frac{3n-2}{n+4}-3} \right\rvert =\left\lvert{\frac{3n-2-3(n+4)}{n+4}}\right\rvert=\frac{14}{n+4}\lt\frac{14}{n}\lt{\epsilon},$$
provided $n\gt\frac{14}{\epsilon}$. Take $n_0\in \mathbb{N}$ with $n_0\geq\frac{14}{\epsilon}$. Then for $n\in \mathbb{N}$ with $n\geq n_0$ we have $|x_n -3|\lt3$, so that $x_n\rightarrow3$ as $n\rightarrow\infty$.
There is no reason why the inequality $\frac{14}n<\varepsilon$ would hold in general. If, for instance, $\varepsilon=n=1$, then it does not hold. Whoever wrote that is actually just saying that$$\frac{14}n<\varepsilon\iff n>\frac{14}\varepsilon,$$which is clearly true.