Except for $x$, all terms are constant and $J \ge 1$.
Been working on this for a while without luck so looking for some help.
Show $Z(x) = \frac{\sinh(2J+1)x}{\ 2J\sinh(\frac{x}{2J})}=2J + 1$ for $x\ll 1$
Attempt:
Let $y = \frac{x}{2J}$
so that we get
$Z(x) = \frac{\sinh(2J+1)(y)}{\sinh(y)}$ but where $x\ll 1$, it is true that $y\ll 1$
so $\sinh(y) = y$
so that $Z(x) = \sinh(2J+1)$.
Any hints would be appreciated.
As noticed in the comments, the expression should be interpreted as follows
$$\sinh\color{red}{\left[(2J+1)x\right]}$$
and then we have indeed, by standard limit $\frac{\sinh t}{t} \to 1$ when $t\to 0$
$$\frac{\sinh\left[(2J+1)x\right]}{\ 2J\sinh(\frac{x}{2J})}=\frac{\sinh\left[(2J+1)x\right]}{(2J+1)x}(2J+1)\frac{\frac x{2J}}{\sinh(\frac{x}{2J})} \to 1 \cdot (2J+1) \cdot 1 = 2J+1$$
which allows to conclude that for $x\ll1$
$$Z(x) \approx 2J + 1$$
For the standard limit we have as $t\to 0$
$$\frac{\sinh t}{t}=\frac{e^t-e^{-t}}{2t}=\frac1{e^{t}}\frac{e^{2t}-1}{2t} \to 1\cdot 1 =1$$