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} $? 

Hint: $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ decreases when $n$ increases to infinity.
