Question: > 
> Find the pointwise limit function of:
>  $$f\_n(x)=\begin{cases} 
>  0 & |x|> 1/n \\
>  nx+1 & x \in [-1/n, 0) \\
>  1-nx & x \in [0, 1/n] 
>  \end{cases} $$
> 
> 
> 

I think that in the limit, if we fix a certain $x$, we get that:

$$\lim \_{n \to \infty} f\_n(x)=\begin{cases} 
 0 & |x|> 0 \\
 \infty & x \in [0, 0) \\
 -\infty & x \in [0, 0] 
 \end{cases} $$
 Where the second line is an empty statement, we rewrite this to:
$$\lim \_{n \to \infty} f\_n(x)=\begin{cases} 
 0 & |x|> 0 \\
 -\infty & x =0 
 \end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$

Answer: 
Your answer is not correct. If $x=0$ then $f\_n(x)=1$ for all $n$ and the limit is $1$. If $x \neq 0$ then $f\_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.
Consider $x\neq 0$. Then there is an $n\_0$ such that $|x|>1/n \hspace{0.3cm}\forall n\geq n\_0 $: thus $f\_n(x)=0$ definitively, and the pointwise limit is 0.

You are left only with $x=0$, but $f\_n(0)=1 \hspace{0.2cm}\forall n$.