$\sin(\alpha) = \frac{\frac{\delta V}{\delta x}}{\sqrt{1 + (\frac{\delta V}{\delta x}})^2}$, $\frac{\delta V}{\delta x}\ll 1$.
$\sin(\alpha)$ can be approximated by:
$\sin(\alpha) \approx \frac{\delta V}{\delta x} + O(\frac{\delta V}{\delta x})^3$.
In this case, why is the additional terms in the order of $O(\frac{\delta V}{\delta x})^3$? i.e. why is the exponent 3 in this case?
since $sin^{'}(\alpha)=cos(\alpha)\alpha^{'}$, $\alpha^{'}$ can be expressed as $\frac{\delta V}{\delta x}$, and $cos(\alpha)=1/sec(\alpha)=1/\sqrt{1+tan^2(\alpha)}$.
since $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-......$, and $(\frac{\delta V}{\delta x})^{4}$'s order is much lower than $(\frac{\delta V}{\delta x})^{3}$. so we can tarnsform $1/\sqrt{1+(\frac{\delta V}{\delta x}})^2$ to $\sqrt{1-(\frac{\delta V}{\delta x})^2}$.
thus the term in your question can be rewrite to $\frac{\delta V}{\delta x}\cdot({\frac{\delta V}{\delta x}})^2$, since $(1-x)^{a}=1+ax+\frac{a(a-1)}{2!}x^2+......$. now your term $O(\frac{\delta V}{\delta x})^3$ holds as a bound $M$ of ratio $[f(x)/(\frac{\delta V}{\delta x})^3]$.