differential geometry - Why is the Taub-NUT instanton singular at $theta=pi$?

Consider the following metric

$$ds^2=V(dx+4m(1-\cos\theta)d\phi)^2+\frac{1}{V}(dr+r^2d\theta^2+r^2\sin^2\theta{}d\phi^2),$$

where

$$V=1+\frac{4m}{r}.$$

That is the Taub-NUT instanton. I have been told that it is singular at $\theta=\pi$ but I don't really see anything wrong with it. So, why is it singular at $\theta=\pi$?

EDIT:: I have just found in this paper that the metric is singular "since the $(1-\cos\theta)$ term in the metric means that a small loop about this axis does not shrink to zero at lenght at $\theta=\pi$" but this is still too obscure for me, any clarification would be much appreciated.