The Question: Determine whether the sequence converges uniformly:
${f_n(x)}: [0,1] \rightarrow \mathbb{R}$,
$f_n(x) = \begin{cases} 0 & f_n(0) = f(2/n) = f(1) = 0 \\ n & x = 1/n \\ \end{cases} $ , and $f$ is linear on the intervals $[0,1/n]$, $[1/n,2/n]$, and $[2/n, 0]$.
The Attempt: Here is a graph of the function: 
There is a typo in the picture. The coordinate $(1/n, 1)$ should be $(1/n,n)$. It was proven that the sequence of functions converges point wise to $f(x) = 0$. It seems obvious to me that the sequence is not uniformly convergent.
Let $N \geq 1$. Choose $\epsilon_0 = 1/3$, Then let $n_0 = N$, and $x= 1/n_0$. Then, $|f_n(x) - f(x)| = |f_n(1/n_0) - f(1/n_0)| = |n_0 - 0| = n0 \geq N > 1/3$. Hence the sequence of functions is not uniformly convergent?
Is this correct or do I need some detail?
Thank you very much!!
What you are doing with $n_0$ and $N$ is a bit confusing: I assume you are trying to negate the definition of convergence to show that the sequence $(s_n)_n$ defined by $$ s_n\stackrel{\rm def}{=} \sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert$$ does not converge to $0$ (and thus that $(f_n)_n$ does not converge uniformly to $f\equiv0$).
In that case, the gist of what you are doing is correct, but it would be, in my opinion, must clear to write the following: let $x_n\stackrel{\rm def}{=} \frac{1}{n} \in[0,1]$ for $n\geq 1$. By definition of the supremum, for any $n\geq 1$ we have $$ s_n \geq \lvert f_n(x_n)-f(x_n)\rvert = \lvert f_n(x_n)\rvert = n $$ and therefore $s_n\xrightarrow[n\to\infty]{} \infty$ (and clearly not $0$).
Now, as a followup question: does it converge uniformly on $(0,1]$? And, more interestingly, on $[a,1]$ for any fixed $a>0$?