Pointwise limit of $f_n(0)=0$, $f_n(1/n)=n$, linear in between What is the pointwise limit of the function $f_n$ which increases linearly on $[0,1/n]$ with
$$
f_n(0)=0,\\
f_n\left(1/n\right)=n,
$$
and is not defined elsewhere, that is, $f_n:[0,1/n] \to \Bbb R$.  
I think the limit is the discontinuous function $f:\{0\} \to \{0,\infty\}$, with
$$
f(0+)=0,\\
f(0-)=\infty,
$$
where
$$
f(0+)=\lim_{x \uparrow 0} f(x).
$$
But this seems strange.
 A: It's clear that $$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty}0=0,$$ so the pointwise limit at $x=0$ is $0$. None of the $f_n$ are defined at negative $x$-values, so the domain of the pointwise limit function will consist only of nonnegative numbers, one of which will obviously be $0$. However, for any $x>0$, there is some $N$ such that $0<\frac1n<x$ for all $n\geq N$--meaning in particular that $f_n(x)$ is undefined for all $n\geq N$, so there isn't a pointwise limit at any positive $x$. Thus (as you've concluded), the domain of the pointwise limit function is $\{0\}$, as is the range (by the work above).
It makes no sense to talk about either $f(0+)$ or $f(0-)$. Neither one of them is defined, since $f(x)$ is undefined for $x\neq 0$.
A: Comment transfered to answer by request.
If you want to talk about pointwise convergence, you have to restrict yourself, to begin with, to the intersection of the domains of your $f_n$'s. 
In this case, this is $\{0\}$.
Now $f_n(0)=0$ for all $n$, so $f_n$ converges pointwise to $0$ on $\{0\}$.
