difficulty to deal with $\min(n,\frac{1}{x})$ in a problem that asks to prove $\{f_n\}_n$ is not uniformly convergent. 
$\{f_n\}_n$ is not uniformly convergent on $[0,1]$, where
$$f_n(x)=\begin{cases} \min(n, \frac{1}{x}) & \textrm{ if } 0<x\leq 1 \\ 0 & \textrm{ if } x=0  \end{cases}.\bigg\}, \space n\in \mathbb{N}$$

I am not sure whether my approach correctly proves that $\{f_n\}$ is not uniformly convergent in $[0,1]$. Specifically, I need to know the way I treated $\min(n,\frac{1}{x})$ whether is correct or wrong.
Here is my approach to prove the problem-
If I'm not wrong then $f_n(x)=\min(n,\frac{1}{x})$ records the lowest value of $n$ and $\frac{1}{x}$ and therefore either $n<\frac{1}{x}$ or $\frac{1}{x}<n$
at $x\to 0, \space f_n(0)=\min(n,\frac{1}{x})\to \infty$ which is meaningless and not sure even if it is pointwise convergent because $$\lim_{n \to\infty}f_n(x)|sup_{x\in(0,1]}\to \infty$$. So there is no sign of uniform convergence or point-wise convergence.
Hence - using cauchy sequence I get, $$|f_n(x)-f(x)|<\epsilon\\\Longrightarrow |\mid n-\frac{1}{x}\mid-\infty|<\epsilon$$ which is absurd iff the assumption $lim_{n\to\infty}\min(n,\frac{1}{x}) \to \infty$ is acceptable. Hence no uniform convergence.
Also there is an explicit formula for which is given as $$\min(x,y)=\frac{1}{2}\Bigg [x+y-|y-x|\Bigg]$$ but using this formula for $\min(n,\frac{1}{x})$ makes the problem more complicated.
Due to the little knowledge on dealing with this kind of problem I ask for any suggestion or any alternate process to prove this. Any help is precious and greatly appreciated.
 A: I think that the term $f_n$ can be rewritten as follows:
$$f_n(x)=\begin{cases} 
n & \textrm{ if } 0<x\leq \frac{1}{n} \\ 
\frac{1}{x} & \textrm{ if } \frac{1}{n} <x\leq 1 \\ 
0 & \textrm{ if } x=0  
\end{cases}.$$
From a point-wise point of view, when $n$ goes to $+\infty$, then the limit of the sequence is:
$$f(x) =\begin{cases} 
\frac{1}{x} & \textrm{ if } 0 <x\leq 1 \\ 
0 & \textrm{ if } x=0  
\end{cases}.$$
Now consider the following:
$$f_n(x) - f(x)=\begin{cases} 
n-\frac{1}{x} & \textrm{ if } 0<x\leq \frac{1}{n} \\ 
0 & \textrm{ if } \frac{1}{n} <x\leq 1 \\ 
0 & \textrm{ if } x=0  
\end{cases}.$$
Then:
$$\sup_{x \in [0, 1]} |f_n(x) - f(x)| = \sup_{x \in \left(0, \frac{1}{n}\right]} \left|n - \frac{1}{x}\right| = +\infty.$$
This means that your sequence does not converge uniformly to the function $f(x)$.
A: The sequence $f_n$ does converge pointwise to
$$
f(x)=\left\{\begin{array}{}
\frac1x&x\in(0,1]\\
0&x=0
\end{array}\right.
$$
This is not continuous, but neither are the $f_n$, so we can't draw a contradiction from this.
If $f_n$ is uniformly convergent, then for any $\epsilon\gt0$, there needs to be an $N$ so that if $n,m\ge N$, then for all $x$,
$$
|f_n(x)-f_m(x)|\le\epsilon
$$
However,
$$
f_{n+1}\!\left(\frac1{n+1}\right)-f_n\!\left(\frac1{n+1}\right)=1
$$
