How to determine pointwise limit/uniform convergence. I have a sequence of functions $f_n:[0,2] \to \mathbb R$ defined by
$$
 f_n(x) =
  \begin{cases}
   n^3x^2
 &  0<x<1/n\\
   n^3(x-\tfrac{2}{n})^2       & 1/n\le x<2/n
\\
   0       & \text{otherwise}
  \end{cases}
$$
I needed to determine the pointwise limit for the function and to determine if $(f_n)$ converges uniformly to $f$. So what I did was:
If $x=0$, $\lim_{n\to \infty} f_n(0)=0$ for all $n$. 
If $x>0$, let $\dfrac{2}{N}<x$, then $f_n(x)=0$ for all $n>N$.
So the pointwise limit of $f(x)$ is $0$. 
But how would I determine if $(f_n)$ converges uniformly? Does it have something to do with the supremum? $\sup|f_n(x)-f(x)|$?
 A: Your reasoning that your sequence  $(f_n)$ converges pointwise to  the zero function $f(x)=0$ on $[0,2]$ is correct.  
Considering uniform convergence now, let's first
recall the definition of uniform convergence of a sequence of functions:

The sequence $(g_n)$ converges uniformly to $g$ on $I$ if for every
  $\epsilon>0$ there is a positive integer $N$ such that $$\tag{1}
  |g_n(x)-g(x)|<\epsilon,\quad\text{for all } n\ge N\ \text{and}\ x\in I. $$  

Note that $(1)$ must hold  for each $n\ge N$ and every $x\in I$. (It does have something to do with the supremum $\sup\limits_{x\in I}|g_n(x)-g(x)|$: this must be be able to be made as small as desired by taking $n$ sufficiently large.)

Back to your sequence.  Note that 
for any positive integer $n$, we have $$\tag{2}f_n(1/n)=n.$$ 
Now let $\epsilon=1 $ and let $N$ be any fixed positive integer. By $(2)$, we have $$|f_N(1/N) -f(1/N)|=f_N(1/N)=N\ge {1 }=\epsilon.$$ 
This shows that for $\epsilon=1$ there is no positive integer $N$ such that $(1)$ holds (with $g_n=f_n$ and $g=f$). Consequently, $(f_n)$ does not converge uniformly on $[0,2]$ (or in fact, on any interval containing $0$).
 
The graphs of the $f_n$ prove illuminating.
The sequence resembles the "witch hat" sequence: it consists of spikes near $x=0$ whose widths approach zero but whose heights grow large. The graphs of the first few terms of the sequence are shown below:
 
