Find the limit function of the given sequence $f_n$. 
For each $n \in \Bbb N$, let 
  $$f_{n}(x) = \begin{cases}nx, &0\leq x\leq \frac{1}{n}\\
1, &\frac{1}{n}<x\leq1\end{cases}$$
  Find the limit function of the sequence $f_{n}$. Show that the convergence of the sequence is not uniform.

I am having difficulty in writing a formal proof. 
As $n$ tends to $\infty$, we have $x=0$, so $f(x)= 0$. So limit function is $f(x)= 0$.
Next step is to show that the convergence is not uniform.
Since for $x=1$ the condition $|f_{n}(x) - f(x)|< \epsilon$ is not satisfied if we take $\epsilon$ less than $1$, so the
convergence is not uniform.
Is this approach correct?
Please guide me to write a formal proof for the first part.
 A: limit function $f$
For $x=0$ holds $x \le \frac1n$ for all $n\in\mathbb N$. So $f_n(0) = 0$ for all $n\in\mathbb N$. Thus $f(0)=0$.
For every $x>0$ exists a $N\in\mathbb N$ such that $x>\frac1n$ for all $n>N$ (Archimedean property). That is $f_n(x) = 1$ for all $n>N$. Thus $f(x)=1$.
So
$$f(x) = \begin{cases}0 & x=0\\1 & 0<x\le 1.\end{cases}$$
no uniform convergence
Proof by contradiction: let be $0<ε<\frac12$. Assume there is a $N\in\mathbb N$ such that for all $x\in[0,1]$ and all $n\ge N$:
$$\vert f_n(x)-f(x)\vert < ε.$$
Choose $x=\frac{1}{2N}$. Then $f_N(x) = \frac12$. So
$$\vert f_N(x)-f(x)\vert = \vert \frac12 - 1\vert = \frac12 \not< ε $$
what is a contradiction.
A: $\lim f_n(0)=0$ is clear. For $x>0$, $|f_n(x)-1|=0<\epsilon $ if $n >\frac 1  x$. So $f_n(x) \to f(x)$ for every $x$ where $f(0)=0$ and $f(x) =1$ for $x >0$.  
If $ 0<\epsilon <1/2$ then $|f_n(\frac 1 {n^{2}}) -f(\frac 1 {n^{2}})|>\epsilon$ for every $n\geq 2$. Hence the convergence is not uniform. 
A: Option.
For $x >0$:
Pick $n_0$ (Archimedean principle) s.t for $n \ge n_0$: $1/n <x$, then $f_n(x)=1$.i.e. $\lim_{n \rightarrow \infty}f_n(x)=1$ for $x>0$.
For $x=0$: $\lim_{n \rightarrow \infty}f_n(0)=0.$.
$f_n$ are continuos on $[0,1]$.
If $f_n$ were uniformly convergent the limit function $f$ would be continuos. Hence?
