Let the sequence of functions $\{f_{n}\}_{n=1}^{\infty}$ be defined by $f_{n}:[0,2] \to \mathbb{R}$ where $f_{n}(x) = \left\{ \begin{eqnarray} 0 &,& 0 \le x < \frac 1 n \\ \frac 1 x &,& \frac 1 n \le x \le 2 \\ \end{eqnarray} \right.$.
I know that the sequence of functions $\{f_n\}$ converges pointwisely to $f(x) = \frac 1 x$. Can someone show that the sequence of functions $\{f_n\}$ does not converge uniformly on $[0,2]$?
Presumably you mean that:
$f_{n}:[0,2] \to \mathbb{R}$ where $f_{n}(x)$ = \begin{cases} \ 0 \textrm{ if $x<\frac{1}{n}$} \\ \ \frac{1}{x} \textrm{ if $x \geq \frac{1}{n}$} \\ \end{cases}
To define uniform convergence: $(\forall \epsilon>0)(\exists N)(\forall x \in [0,2])(n \geq N \implies |f_{n}(x)-f(x)<\epsilon)$
Therefore to negate this: $(\exists \epsilon)(\forall N)( \exists x \in [0,2])(n \geq N \implies |f_{n}(x)-f(x)|>\epsilon$.
To show this is not uniformly continuous, let $\epsilon=1$ Let $N \in \mathbb{N}$. Let $x=\frac{1}{N+1}$
Then $x<1/N$ for each $N \in \mathbb{N}$ So $|f_{n}(x)-f(x)|=|0-N+1|=|N+1|>1=\epsilon$.
Therefore, we found an epsilon such that for each choice of $N$, there exists some $x$ such that $f_{n}(x)-f(x)|>\epsilon$ and we are done.
To show pointwise convergence [I know you said you got this, but I want to hopefully make clear the difference]: you want
$(\forall x \in [0,2])(\forall \epsilon>0)(\exists N \in \mathbb{N})(n \geq N \implies |f_{n}(x)-f(x)<\epsilon)$
The order of the quantifiers makes the whole difference.
Let $x \in [0,2]$. Let $\epsilon>0$. Then we want to somehow arrive at $|f_{n}(x)-f(x)|<\epsilon$.
Well, by the archimedean property, there exists some $N \in \mathbb{N}$ such that $\frac{1}{N}<x$, meaning that if $n>N$, we have that $\frac{1}{n}<\frac{1}{N}<x$. But then $f_{n}(x)=\frac{1}{x}$, trivially implying that $|f_{n}(x)-f(x)|=0<\epsilon$.