Chapter 3.6, Problem 4E

In Example 1, page 126, the improved Euler’s method approximation to $e$ with step size $h$ was shown to be

${\left(1+h+\frac{h^2}{2}\right)}^{1/h}$.

First prove that the $\text{error}:=e-{\left(1+h+h^2/2\right)}^{1/h}$ approaches zero as $h\to 0$. Then use L’Hôpital’s rule to show that

$\underset{h\to 0}{\lim }\frac{\text{error}}{h^2}=\frac{e}{6}\approx 0.45305$.

Compare this constant with the entries in the last column of Table 3.5.