<s>[INST] Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $ Prove $\lim_{n \rightarrow  \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0   $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} - 0  \right | < \epsilon$.
$\Leftrightarrow \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$ 
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon \Leftrightarrow n > \frac{4}{3} \cdot \frac{1}{\epsilon} \Rightarrow n \geq N$
Take $N = \left \lfloor \frac{4}{3} \cdot \frac{1}{\epsilon} \right \rfloor + 1$
In the above proof, I'm confused as to how $\frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$ 
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon$. How did $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ become $\frac{4}{3} $? 
 [/INST] Since $n \in \mathbb N \implies n \geq 1$, we know that:
$$
\frac{1}{n} \leq 1 \iff 3 + \frac{1}{n} \leq 4
$$
as well as that:
$$
\frac{-1}{n} \geq -1 \iff 4 - \frac{1}{n} \geq 3
$$
Now to make a fraction smaller, we can either decrease the numerator or increase the denominator. Thus, it follows that:
$$
\epsilon > \frac{1}{n} \cdot \frac{4}{3}
\geq \frac{1}{n} \cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}
$$
</s>