Let $u:[0,\infty)\to\mathbb R$ be a continuous function such that $\lim_{x\to\infty} u(x)$ exists and is finite. Show that one can approximate $u(x)$ uniformly by linear combinations of $e^{nx}$.
It was not mentioned if $n$ is merely an integer or a positive integer. I can do it only for negative $n$. Note, $x>0$.
My argument is based on getting the $k$-th Bernstein polynomial $B_k(x)$ for $v(x):=u(-\log(x))$ on $[0,1]$ and thus $B_k(e^{-x})$, which is a finite linear combination of $1, e^{-x},...,e^{-kx}$ will approximate uniformly $u(x)$.
Since $e^{-x/n}=1-x/n+(x/n)^2/2+\cdots$, you can approximate $1$ uniformly by $e^{-x/n}$ by large enough $n$ on on any finite interval. Similarly, $x\approx n(1-e^{-x/n})$ for large enough $n$. This gives you approximations for $x^a\approx n^a(1-e^{-x/n})^a$. It follows that you can approximate polynomials uniformly on any finite interval via linear combinations of $e^{-x/n}$. Then if $L=\lim_{x\rightarrow\infty}u(x)$, you can approximate $u(x)-L$ via polynomials on any finite interval, and therefore by linear combinations of $e^{-x/n}$. If you take $x$ large enough, all the exponential terms die, and since you can uniformly bound $|u(x)-L|<\epsilon$ for all $x>N$, that should seal the deal.