Let $R = \frac{g_{J}\mu_{B}B_{ext}}{K_{B}T}, x = \frac{R}{g_{J}}, B_{J}(x) = \frac{2J+1}{2J}coth(\frac{(2J+1)x}{2J}) - \frac{1}{2J}coth(\frac{x}{2J})$
In the limit of small $B_{ext} \space or \space R<<1:$ show that $B_{J} \approx \frac{(J+1)R}{3}$
Attempt:
$lim_{B_{ext} \rightarrow 0}R(B_{ext}) \rightarrow lim_{R \rightarrow 0} x(R)$ so $B_{ext}$ results in $x \rightarrow 0$
$lim_{x \rightarrow 0}Coth(x) = lim_{x \rightarrow 0} \frac{cosh(x)}{sinh(x)} \approx \frac{1}{x}$
This gives (substitute the form for $x$):
$B_{J}(x) = \frac{2J+1}{2J} (\frac{2JK_{B}T}{(2J+1)\mu_{B}B_{ext}}) - (\frac{1}{2J})(\frac{K_{B}T}{\mu_{B}B_{ext}}) = 0$
which fails to coincide with the required claim.
Any help to illuminate my errors are appreciated.
The problem is so uncompletely specified (we don't know which symbols denote constants and which are variables, and who depends on what, and how) that the only thing that can be said about it is the following.
Define $$f_\alpha(x)=\alpha\coth(\alpha x).$$ Then, as $x\to0$ (with $\alpha$ fixed) $$\begin{aligned}f_\alpha(x)&=\frac\alpha{\tanh(\alpha x)}\\&=\frac\alpha{\alpha x-(\alpha x)^3/3+o(x^3)}\\&=\frac1{x\left(1-(\alpha x)^2/3+o(x^2)\right)}\\&=\frac{1+(\alpha x)^2/3+o(x^2)}x\end{aligned}$$ hence (for $\alpha,\beta$ fixed and distinct) $$f_\alpha(x)-f_\beta(x)\sim\frac{\left(\alpha^2-\beta^2\right)x}3.$$ Now, $$\left(\frac{2J+1}{2J}\right)^2-\left(\frac1{2J}\right)^2=\frac{J+1}J$$ hence as $x\to0$ (for $J$ fixed) $$\frac{2J+1}{2J}\coth\left(\frac{(2J+1)x}{2J}\right)-\frac1{2J}\coth\left(\frac x{2J}\right)\sim\frac{(J+1)x}{3J}.$$