Solution
= Solution
The definition of <hyperbolic sine> simplifies the composition exactly for $x>0$:
$$
\sinh(\ln x)=\frac12\left(e^{\ln x}-e^{-\ln x}\right)=\frac12(x-x^{-1}).
$$
Set $h=x-1$. The <geometric series> $1/(1+h)=1-h+h^2-h^3+h^4+O(h^5)$, valid for $|h|<1$, then gives
$$
\boxed{\sinh(\ln x)=(x-1)-\frac12(x-1)^2+\frac12(x-1)^3-\frac12(x-1)^4+O((x-1)^5).}
$$