Solution
= Solution
The linear terms cancel: $\tan x=x+x^3/3+O(x^5)$, whereas $\tanh x=x-x^3/3+O(x^5)$. Also $\sinh x-x=x^3/6+O(x^5)$. Their leading nonzero terms therefore give
$$
\boxed{\lim_{x\to0}\frac{\tan x-\tanh x}{\sinh x-x}=\frac{2/3}{1/6}=4.}
$$
The <Maclaurin series> remainder orders justify division after factoring out $x^3$.