Trouble in finding limit of a sequence of functions

Question

Let's consider the sequence of functions $\{f_n(x)\}$ defined as follows:
$f_n(x) = \begin{cases} 2n^2x, & \text{$0\le x \le \frac{1}{2n}$} \\ 2n-2n^2x, & \text{$\frac{1}{2n}\le x\le \frac{1}{n}$}\\ 0, & \frac{1}{n}\le x\le 1 \end{cases} \tag{1}$
Intuitively, it seems from (1) that as $n \to \infty, 1/n, 1/2n $ tend to $0$ hence eventually for large enough n, we end up with $f_n(x)=0$ for all $x \gt 0.$ I am trying to write a formal proof of it.
So I have to show that $\forall \epsilon \gt 0, \exists N$ such that for all $n \ge N$, $|f_n(x)-0|\lt \epsilon \tag{2}$
(2) should be correct for every definition of $f_n(x)$ in (1). If I take $f_n(x)=2n^2 x $ then clearly $|f_n(x)|=2n^2x\le n$, which can not be bounded by $\epsilon \implies \lim_{n\to \infty} f_n(x)$ does not exist. But the limit actually exists. How can I write a formal proof for $\lim_{n\to \infty} f_n(x)=0$?

Answer 1

For $ x=0$,

$$f_n(0)=0 \implies \lim_{n\to+\infty}f_n(0)=0=f(0)$$

For $x>0$ and $n $ great enough, $$n\ge \lfloor \frac 1x \rfloor +1,$$ we will have $$\frac 1n < x$$ and $$f_n(x)=0$$

So,

Given $x\in (0,1] $ and $ \epsilon>0$

there exists $ N=\lfloor \frac 1x \rfloor +1 $ such that

$$(\forall n\ge N )\;\;\; |f_n(x)-0|=0<\epsilon$$

Thus $\lim_{n\to+\infty}f_n(x)=0$.

We conclude that

the sequence is pointwise convergent to the zero function $ f $ at $[0,1]$.

Answer 2

The given formulas obscure what's going on. It's really easier than that. Here it is in a general form: Suppose $f_n$ is a sequence of functions on $[0,1]$ such that i)$f_n(0)=0$ for all $n$ and ii) for each $n,$ $f_n=0$ on $[1/n,1].$ Then $f_n\to 0$ pointwise everywhere on $[0,1].$