Consider a conical pendulum. Let $\alpha$ be the acute angle between the string and the vertical, $T$ be the magnitude of the tension in the string and $mg$ be the magnitude of the gravitational force acting on the pendulum vertically downwards.
Then, according to my 1st year Mathematics BSc Vectors and Mechanics course, \begin{align} T\cos\alpha=mg,\tag{1} \end{align}
so \begin{align} T=\frac{mg}{\cos\alpha}.\tag{2} \end{align}
From this, it seems to me, it follows that \begin{align} \lim_{\alpha\rightarrow\frac{\pi}{2}}T&=\lim_{\cos\alpha\rightarrow 0}T\tag{3.1}\\ &=\frac{mg}{0}\tag{3.2}\\ &=\infty\tag{3.3}, \end{align}
but this is unphysical, since the string of a horizontally spinning pendulum does not have infinite tension.
What have I missed?
This is physical. No pendulum on earth will ever spin completely horizontal--it will always sag a bit under gravity. In order to spin truly horizontal, it would have to spin infinitely fast, and would then require infinite string tension.