How can I prove this sequence of functions pointwisely converges to zero using formal argument?

Question

Let $f_n(x)$ be a sequence of functions defined on $ [0,1]$ by

$$\ f_n(x)= \begin{cases} n \text { if $0< x < \frac{1}{n}$}, \\ 0 \text { if $x=0$ or $ \frac{1}{n}\le x \le 1$ }\end{cases}$$

How to prove the pointwise limit of $f_n(x)$ is $0$ using $\epsilon-\delta $ argument ?

Answer 1

Hint:

Fix $x \in (0,1]$ and pick $n_0 \in \mathbb{N}$ such that $\frac1{n_0} \le x$. Then $f_n(x) = 0$ for all $n \ge n_0$.

Answer 2

$\textbf{Solution}$: Fix $x \in (0,1)$, as if $x=1$, then $f_n(1)=0$ for every $n$ and we have the claim and similarly for $x=0$. So now assume WLOG $x \in (0,1)$, and let $\epsilon > 0$. Then by archmedian's principle, there exists an $N \in \mathbb{N}$ such that \begin{align} \frac{1}{N} < x \end{align} In particular, $x \in (1/N,1) \subset [1/N,1]$. Then \begin{align} f_N(x) = 0 \end{align} Then we notice that for any $n \geq N$, $\frac{1}{N} \geq \frac{1}{n}$, so we have $(1/N,1) \subset (1/n,1)$. In particular, this means for any $n \geq N, f_n(x) = 0$. Then the rest follows from $0 < \epsilon$.