The Nernst equation can be applied to the whole cell reaction or a half-cell reaction. Consider the half cell reaction: $$\ce{Cu^{2+} + 2e^- -> Cu(s) \quad \quad E^0 = +0.337~\rm{V}}$$
The Nernst equation expressed in base 10 log at 25 °C is:
$$E = E^0 - \frac{0.1596}{2}\log_{10} \frac{a_{\ce{Cu^2+}}}{a_{\ce{Cu}}}$$
But $a_{\ce{Cu}} = 1$ by definition, and for low concentrations of $\ce{Cu^2+}$ where
$a_{\ce{Cu^{2+}}} \ll 1$ then: $\ce{[Cu^2+]} \approx a_{\ce{Cu^2+}}$.
But when $\ce{[Cu^2+]} < 1$ then the log will be negative so:
$$\lim_{\ce{Cu^2+}\to 0} (0.337 - 0.0798 \times \log_{10} [\ce{Cu^2+}]) = \infty $$
This doesn't seem realistic. For instance $\ce{[Cu^2+] = 1.0\times 10^{-1,000,000}}$ is fine mathematically, but that sort of concentration doesn't make any sense chemically.
Thus what is the limit of the half cell potential as the $\ce{[Cu^2+] -> 0}$?
Edit-1 - My gut reaction to this is that you can't really measure the activity of a single ion. If the ion of interest is low in concentration compared to the ionic strength in general, then it would seem that the activity of that ion should go to zero as its effect on the total overall ionic strength goes to zero.