<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] Observe that for all $n \geq 1$, $3 + \dfrac{1}{n} \leq 4$, and $4 - \dfrac{1}{n} \geq 3$. Thus: $\dfrac{3+\dfrac{1}{n}}{4 - \dfrac{1}{n}} \leq \dfrac{4}{3}$. Thus given $\epsilon > 0$, choose $N_{0} =\lfloor\dfrac{4}{3\epsilon}\rfloor + 1$, then if $n > N_0$, we have: $\dfrac{1}{n}\cdot \dfrac{3+\dfrac{1}{n}}{4 - \dfrac{1}{n}} < \dfrac{4}{3n} < \epsilon$, and the conclusion follows.
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