$f_n(x)$ converges pointwise to $f(x)$ If the function $f_n(x)$ converges pointwise to $f(x)$, where:
$$f_n(x) = \frac{1}{\exp(2nx)}$$ for $$x\in[0,1]$$
Would $f(x)=0$?
My initial thoughts were to separate $f_n(x)$ into two limits and add them back together afterwards. Is this a valid method? 

Rewrite $f_n(x)$ to
$$
f_n(x)=\left(e^x\right)^{-2n}
$$
For $x=0$ we have $e^x=1$ and $f_n(0)=1^{-2n}=1$
For $x>0$ we have $e^x>1$ and therefore $f_n(x)\to0$ as $n\to\infty$.
As such we get that
$$
f(x)=\begin{cases}1&\text{if }x=0\\0&\text{if }0<x\le1\end{cases}
$$
As Arthur correctly points out in the comments, a sequence of continuous functions converging pointwise need not converge to anything continuous.
