Our total error becomes $$\left|\epsilon_{\mbox{tot}}\right|\le \frac{2 \epsilon_M}{h^2} + \frac{f_0^{(4)}}{12}h^{2}.$$ It is then natural to ask which value of $h$ yields the smallest total error. Taking the derivative of $\left|\epsilon_{\mbox{tot}}\right|$ with respect to $h$ results in $$h= \left(\frac{ 24\epsilon_M}{f_0^{(4)}}\right)^{1/4}.$$ With double precision and $x=10$ we obtain $$h\approx 10^{-4}.$$ Beyond this value, it is essentially the loss of numerical precision which takes over.