Finidng the infinity limit of $\coth$ function. I have the function:
$$ g(x) = \lim_{J \to +\infty} \frac{1}{2J} \coth(\frac{x}{2J}) $$ 
In the answers it gives:
$$ g(x) = \frac{1}{2J}\frac{2J}{x} $$.
I don't understand how the infinity limit of the coth function was found. Any help would be appreciated, thank you! 
Edit:
Part of the larger question:
$$ L(x) = \lim_{J \to +\infty} \Big[ \frac{1}{J} f_{2J+1} \Big( \frac{x}{J} \Big) \Big], $$
where the Brilllouin function, $f_n(x)$, is defined by:
$$ f_n(x) = \frac{n}{2} \coth \Big( \frac{nx}{2} \Big) - \frac{1}{2} \coth \Big( \frac{x}{2} \Big). $$ 
And substituting this in gives: 
$$ L(x) = \lim_{J \to +\infty} \frac{1}{J} \Big[ (J + \frac{1}{2}) \coth \Big[(1 + \frac{1}{2J})x \Big] - \frac{1}{2}\coth(\frac{x}{2J}) \Big]  $$
 A: For small $y$, $\coth y\sim\frac1y$. As $J\to\infty$, $\frac{x}{2J}\to0$ so $\coth\frac{x}{2J}\sim\frac{1}{\frac{x}{2J}}=\frac{2J}{x}$. Hence $\frac{1}{2J}\coth\frac{x}{2J}\sim\frac1x$.
A: For small $x$, $\operatorname{sinh}(x) \cong x$ and $\operatorname{cosh}(x) \cong 1$ by Taylor-expansion up to first order. Thus, $\operatorname{coth}(x) = \operatorname{cosh}(x)/\operatorname{sinh}(x)\cong 1/x$. One does not account for higher orders as in the limit of big $x$, they can easily be seen to drop out of the limit we want to take (to be mathematically precise, a more elaborate discussion would be needed here, but I see that you are doing physics so this should work out).
With that, you should easily see what happens here.
A: Here is another way to see that $\coth x \sim \frac{1}{x}$ when $x$ is small:
$$\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$$
For small $x$, $e^x \approx 1 + x$ from the first two terms of its Taylor series. Therefore:
$$\coth x \approx \frac{1+x+1-x}{1+x-(1-x)} = \frac{2}{2x} = \frac{1}{x}$$
A: Regarding your "simple" version of $L(x)$, @J.G. and others already gave you all the details. Now apply these rules to find (for the full expression of $L$),
$$
L(x) = \lim_{J \to +\infty} \frac{1}{J} \Big[ (J + \frac{1}{2}) \coth \Big[(1 + \frac{1}{2J})x \Big] - \frac{1}{2}\coth(\frac{x}{2J}) \Big] = \coth(x)-\frac{1}{x} 
$$
