Temperature limit on entropy of a paramagnet We have
$$S=Nk_B[\ln(2 \cosh(x)) - x \tanh(x)]$$ where $$x = \frac{\mu B}{k_BT}$$
In need to show that at low temperatures entropy
$$S \approx Nk_B2xe^{-2x}$$
I wrote out the $\cosh(x)$ in terms of $e^{-x}$ and get that $\ln(2\cosh(x))\approx e^{-2x}$ at low temperatures, which would then mean I need to way to write $x \tanh(x) \approx e^{-2x}-2xe^{-2x}$, which I cannot seem to find
 A: Let me expand on the previous answer. We expand each term separately. We have,
\begin{align} 
\log \left( 2 \cosh x  \right)  & = \log \left( e ^x + e ^{ - x } \right) \\ 
& = \log \left[ e ^{ x } ( 1 + e ^{ - 2 x } ) \right] \\
& = x + \log \left( 1 + e ^{ - 2 x } \right) \\ 
& \approx x + e^{-2x}
\end{align}
For the second term we have,
\begin{align} 
x \tanh ( x ) & = x \frac{ e ^{ x } - e ^{ -  x } }{ e ^{ x } + e ^{ -  x }  } \\ 
& = x \frac{ 1 - e ^{ - 2 x } } { 1 +  e ^{ - 2x } } \\ 
& \approx  x ( 1 - e ^{ - 2  x } ) ^2  \\ 
& \approx x ( 1 - 2 e ^{ - 2 x } ) 
\end{align}
where all corrections are of order $ (xe ^{ - 2 x } ) ^2 $. Notice that $xe^{-2x}$ in the large $x$ limit is much greater then $e^{-2x}$ since $x $ is large. Thus to next to leading order we can drop the $e ^{-2x}$ in the hyperbolic $\log(\cosh x)$ approximation.
Thus in total we have,
\begin{equation} 
S \approx 2N k _B x e ^{ - 2 x } 
\end{equation}
A: They probably mean to say "for large $x \gg 1$". The limit of low temperature is vague if they don't say how quantities are being fixed in the limit. Anyway, your approximations are wrong. For large $x$, $\ln(2 \cosh x)$ goes like $x$, and $\tanh x$ goes like $1 - 2 e^{-2x}$. Try graphing it if you don't believe me. :)
