I'm reading notes about spectral theory and at some point it is stated that the $\mathbb{C}$-linear span of the functions $f_{\lambda}:\mathbb{R}\to\mathbb{C}:x\mapsto (x-\lambda)^{-1}$, where $\lambda\in\mathbb{C}\setminus\mathbb{R}$, is dense in the space $C_{0}(\mathbb{R};\mathbb{C})$ of continous functions vanishing at infinity, for the supremum norm $\Vert\cdot\Vert_{\infty}$. It should be a consequence of the (locally compact version of) the Stone-Weierstrass theorem. I therefore write $A:=\text{span}\left(\{f_{\lambda} : \lambda\in\mathbb{C}\setminus\mathbb{R}\}\right)$, and try to prove that $A$ is a subalgebra of $C_{0}(\mathbb{R};\mathbb{C})$.
If $\lambda\neq\mu$, then by partial fraction decomposition we have $$f_{\lambda}f_{\mu}=(\lambda-\mu)^{-1}f_{\lambda} + (\mu-\lambda)^{-1}f_{\mu}\in A.$$ However, when $\lambda=\mu$, how do I decompose $x\mapsto f_{\lambda}(x)^{2}=(x-\lambda)^{-2}$ as an element of $A$ ?
It is not true that $f_\lambda^2\in A$. However, $f_\lambda^2$ is in the closure of $A$, which is good enough to conclude that the closure is a subalgebra and thus all of $C_0(\mathbb{R};\mathbb{C})$ by Stone-Weierstrass. To prove this, just note that $f_\lambda^2$ is uniformly approximated by $f_\lambda f_\mu$ as $\mu$ approaches $\lambda$.