For $n \in \mathbb{N}$ let's consider $f_{n}: \mathbb{R}_{+} \to \mathbb{R}, x \mapsto e^{-nx}$, and let $E = \text{span}(f_{n})_{n \in \mathbb{N}}$.
The question is: determine the functions that are uniform limits of sequences of $E$.
So obviously such functions are continuous, but else ... I do not know ... Maybe the fact that $f_{n} = (f_{1})^{n}$ is useful, but I can't see how.
They are precisely continuous functions $f$ such that $f(x) \to 0$ as $x \to \infty$.
Proof: Given any such function $f$ define $g: [0,1] \to \mathbb R$ by $g(x)=f(- \ln x)$ if $x >0$ and $g(0)=0$. You can check that $g$ is continuous. By Weierstrass Theorem there is a polynomial $p$ such that $|g(x)-p(x)| <\epsilon$ for all $x$. Note that $|p(0)| <\epsilon$ since $g(0)=0$. We may therfore modify $p$ to a polynomial without constant term. This translates to $|f(x)- \sum\limits_{k=1}^{n} c_ke^{-kx}|<\epsilon$.