Having trouble showing continuity, and pointwise limit of this function. The function that I am having trouble with is: 
$$ { f }_{ n }(x)=\begin{cases} 1\quad ,\quad x \in\{ 1,\frac { 1 }{ 2 } ,...,\frac { 1 }{ n } \}  \\ 0\quad ,\quad otherwise \end{cases} $$ 
and in particular I am trying to show that the each $f_{n}$ is continuous at x = 0, and also I am trying to find the pointwise limit of this function.


*

*For the continuity at x=0 I believe I would have to claim that since the set:
$$ {1,\frac{1}{2},...,\frac{1}{n} } $$ tends to zero as $ n \rightarrow \infty $ and that the second condition would be when x is exactly 0 it is zero because it doesn't belong in the set so it would then be continuous at x=0. 

*As for the pointwise I belive I would use the same argument to show that the pointwise limit would be 0. 


If these are correct or incorrect any suggestions would be appreciated.
 A: Indeed, it is true that $f_n$ is continuous at $x=0$. Note that, $f_n(x)=0$ for $x\in(-\frac1n, \frac1n)$, that is, $f_n$ is constant in a neighbourhood of $x=0$.
Now, for the pointwise limit, note that if $m\in\mathbb N$, then $f_n(\frac1m)=1$ if $n\geq m$. For the other points, it is $f_n(x)=0 \ \forall n\in\mathbb N$. 
A: The continuity of $f_n(x)$ in $0$ is ensured because this function is constant in a neighbourhood of $0$ i.e. $\forall |x|<\frac{1}{n},\ f_n(x)=0$.
The pointwise limit is defined by 
$f(x)=\begin{cases}
1 \quad \forall x\in\{\frac 1k\,|\,k\in \mathbb N^*\}\\
0 \quad otherwise\\
\end{cases}$
This function $f$ is not the null function, there are an infinity of points where its value is $1$.
But $f$ is not continuous in $0$, because for any $\delta>0$, let's take $n_0=\lfloor \frac 1\delta\rfloor+1$ so that $\frac 1{n_0}<\delta$
$f$ continuity would be $\forall \varepsilon>0,\ |x|<\delta\Rightarrow|f(x)<\varepsilon$ but $f(\frac 1{n_0})=1>\varepsilon$ so this is failing.
