Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $

Question

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} $?

Answer 1

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

Answer 2

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.

Answer 3

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: $$ \epsilon > \frac{1}{n} \cdot \frac{4}{3} \geq \frac{1}{n} \cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} $$