What is the sideways tension on a hanging chain under gravity? I'm sure you are familiar with the question of deducing the shape of a free hanging chain but if not, here's my solution so far summarised:

By taking $\tan(\theta)$ to be $f'(x)$ and considering the forces on a
  general particle, one can show that the shape of a free hanging chain
  obeys $f''^2(x) = (1+f'^2(x))k^2$ whose solution is
  $\frac{1}{k}\cosh(kx)$ and $k = \frac{g\rho}{T_0}$ where g is
  gravitational acceleration, $\rho$ is length-density $(pL = m)$ and $T_0$ is
  sideways tension (which must remain constant otherwise particles in
  the chain would accelerate sideways). The general solution to the diff
  equation is also works for sideways and vertical shifts of this
  function but these can be set to zero.

Now my problem is finding $T_0$ given initial conditions. Note that the differential equation is not affected by changing the distance between the two endpoints (call it $2x_0$) so that is also an initial condition.
I tried looking at the endpoint of the chain where $2T_0 = \frac{mg}{\sinh(kx_0)}$ but this leads to an equation which gives a nonsensical answer when $kx_0$ approaches zero. If $kx_0$ approaches zero then $2T_0 = \frac{mgx_0}{gm/(LT_0)}$ so $x_0 = L/2$, which seems nonsensical. Maybe this is because $T_0$ becomes small with $x_0$, preventing the $kx_0$ approaching zero? A problem I'm having is that there are no real numerical solutions when I plug this function into wolfram alpha. Please help me
 A: The answer on Physics SE, linked by cgiovanardi in a comment, considers a non-symmetric configuration, where the two ends are at different heights. If you are only interested in the symmetric case, it is somewhat simpler than that, though you can still apply the same approach, and in the end the equation is transcendental anyway.
The usual method is to consider the length of the chain, which is also an initial condition. From the length, together with distance between endpoints, you can get the $k$ - that's just the geometry of the curve. Then $k$ tells you what $T_0$ is. 
Your approach doesn't do that explicitly, but still gets the same result because $m = \rho L$, so you end up with the same equation (though you have a subtle mistake in that the coordinates of the endpoints are $\pm x_0/2$ with your definition of $x_0$):
$$ \sinh\frac{kx_0}2 = \frac{kL}2 \;\;\;\Leftrightarrow\;\;\; \frac{\sinh(kx_0/2)}{kx_0/2} = \frac{L}{x_0} $$
Notice that $x_0 \rightarrow L \Rightarrow kx_0 \rightarrow 0$ but it still makes perfect sense - that is a too-tightly-stretched chain with $k \rightarrow 0 \Rightarrow T_0 \rightarrow \infty$; while as $x_0 \rightarrow 0$, you just get a vertically hanging doubled-up chain, which will not have a horizontal component of tension: $kx_0 \rightarrow \infty \Rightarrow k \rightarrow \infty \Rightarrow  T_0 \rightarrow 0$.
