= Solution
Suppose $b$ were a <left inverse of an algebra element> $x$, so $bx=1$. Then $bx_n\to1$, and $bx_n$ is invertible for all sufficiently large $n$ by the <Neumann series>. For such $n$, $b=(bx_n)x_n^{-1}$ is a product of invertibles, hence invertible. The equation $bx=1$ then gives $x=b^{-1}$, a contradiction.
If instead $xb=1$, then $x_nb\to1$, so $x_nb$ is eventually invertible. Now $b=x_n^{-1}(x_nb)$ is invertible, and again $x=b^{-1}$, a contradiction. This proves that \b[a noninvertible limit of invertibles has neither type of one-sided inverse]. No finite-dimensional assumption is used.
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