Solution
= Solution
Use the <Taylor series> $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the <Taylor series> is
$$
\boxed{f(x)=1+x^3+O(x^5).}
$$
The linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in <big O notation>, follows because the fourth-order term is also zero; the next term is $-x^5$.