Finding the pointwise limit of a sequence of elementary functions defined by n * the characteristic function. I basically want to find the limit of the following sequence:
$f_n = nX_{[0, 1/n]}$
$X := 1$ if $x \in [0, 1/n], x = 0$ otherwise.
When I look at this function, my first thought is that the limit approaches infinity and therefore doesn't exist, since even though x would have to be closer and closer to 0 for the function to not equal 0, n*1 = n.
However, according to the problem description, a limit f should exist. I'm afraid my understanding of this topic is still very basic so if someone could help explain, I'd appreciate it.
 A: The limit is the function $$f(x)=\left\{\begin{array}{ll}\infty & : x = 0 \\ 0 & : x\neq 0.\end{array}\right.$$
You say "the limit approaches infinity," but the limit should be a function, not a number. However, the function can be $\infty$-valued. In this case it is, at $0$.
To see why, we note that if $x<0$, then $f_n(x)=0$ for all $n$, so $\lim_n f_n(x)=\lim_n 0=0$.
If $x=0$, then $f_n(x)=f_n(0)=n$, and $f(0)=\lim_n f_n(0)=\lim_n n=\infty$.
For $x>0$, there exists $n_0\in\mathbb{N}$ such that $1/n_0<x$. Then $f_n(x)=0$ for all $n\geqslant n_0$. So $\lim_n f_n(x)=0$.
A: The answer will depend on what sense you wish to find a limit. In the pointwise sense, there is no function $f(x) = \lim_n f_n(x)$ because of the issues at $0$. However, if you only care about almost everywhere convergence, then it converges to the zero function. If you allow 'generalized functions', i.e. distributions, then $\lim_n f(n) = \delta_0(x)$, the Dirac delta at zero. There are other notions of limits that may produce different results but these are the two mains one I think.
