Solution
= Solution
Use the <exponential series> to obtain the <Maclaurin series> of the <hyperbolic sine> and <hyperbolic cosine>:
$$
\sinh x=x+\frac{x^3}{6}+O(x^5),\qquad \cosh x=1+\frac{x^2}{2}+O(x^4).
$$
Dividing these <power series> gives the <hyperbolic tangent> expansion
$$
\tanh x=\left(x+\frac{x^3}{6}+O(x^5)\right)\left(1-\frac{x^2}{2}+O(x^4)\right)=\boxed{x-\frac{x^3}{3}+O(x^5)}.
$$
The displayed remainder uses <Big O notation> as $x\to0$.