Pointwise limit function of a piecewise function

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 1

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 $0$.

Answer 2

Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $x\neq 0$?) But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.

Answer 3

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 $f_n(0)=1 \hspace{0.2cm}\forall n$.