# Show $\lim _{n\rightarrow \infty }\frac {5} {n}=0$

My proof:

Let $\varepsilon >0$. By the Archimedean property, there is a $N$ in $\mathbb{N}$ such that $\dfrac {1} {N}\varepsilon < 1$. Hence, we obtain $N < \dfrac {1} {\varepsilon }$. Thus, for all $n\geq N$ $$\left| \dfrac {5} {n}-0\right|=\dfrac {5} {n}=\dfrac {1} {n(\dfrac {1} {5})}<\dfrac {1} {n}<\dfrac {1} {N}<\varepsilon.$$ We are done.

Can you check my proof and mathematical writing (proof-writing)?

• It is not true that $\frac{5}{n} < \frac{1}{n}$. Perhaps just proceed from $|\frac{5}{n}| = \frac{5}{n} \leq \frac{5}{N}$ since $n\geq N$ and choose $N$ such that $\frac{5}{\epsilon} < N$. – Weaam Aug 10 '17 at 23:02
• @Weaam Thanks.. – James Ensor Aug 11 '17 at 0:10

As Weaam wrote, $\dfrac {1} {n(\dfrac {1} {5})} <\dfrac {1} {n}$ is false.
You want $n$ to be large enough so that $\dfrac {5} {n} <\varepsilon$. This is equivalent to $\dfrac {5} {\varepsilon} <n$, and this is the lower bound on $n$ that is needed.
• Then, we should say 'by the Archimedean property there is a $N$ in $\mathbb{N}$ such that $N\varepsilon > 5$. Hence, we obtain $\dfrac {5} {N } < \varepsilon$. – James Ensor Aug 11 '17 at 0:16
Let $ε>0$ By the Archimedean property,$\quad\exists n_0\in\mathbb{N},\quad n_0\varepsilon>5\iff n_0>\dfrac{5}{\varepsilon}$
We can choose $n_0=\left\lfloor\dfrac{5}{\varepsilon}\right\rfloor+1$