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 $n_0\gt ...$ ".


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.


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.

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    $\begingroup$ OMG!Our answers are actually the same. Even the example!! $\endgroup$ – Thomas Shelby Feb 14 at 18:04
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    $\begingroup$ @ThomasShelby Well, after all it was the natural example in this context. $\endgroup$ – José Carlos Santos Feb 14 at 18:06
  • $\begingroup$ So would it be ok to proceed with "take $n_0 \in \mathbb{N}$ with $n_0 \gt \frac{14}{\epsilon}$ so that for $n\in \mathbb{N}$ with $n\geq{n_0}$ we have that $|x_n - 3|\lt{\epsilon}$. Therefore $x_n\rightarrow{3}$ as $n\rightarrow{\infty}$" $\endgroup$ – Adnaan Feb 14 at 18:07
  • $\begingroup$ No, it would not be OK. Pick $n_0$ such that $n_0>\frac{14}\varepsilon$ instead. $\endgroup$ – José Carlos Santos Feb 14 at 18:08

Let's take $\varepsilon =1$. Then note that $\frac {14}1\gt1$.So we don't have $\frac{14}n\lt 1$ for all $n\in \Bbb N $. But, if you take $n\gt 14$, we have $\frac {14}n\lt1$. So in general, $$\frac {14}n\lt\varepsilon \iff n\gt\frac {14}{\varepsilon}$$.


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