Let's do some calculus! (28)

$\begin{aligned} L & = \lim_{x \to 0} \frac {\cos^2 x - \cos x - e^x \cos x + e^x - \frac {x^3}2}{x^n} \\ & = \lim_{x \to 0} \frac {(\color{#3D99F6}{\cos x} - \color{#D61F06}{e^x})(\color{#3D99F6}{\cos x} - 1) - \frac {x^3}2}{x^n} & \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \lim_{x \to 0} \frac {\left(\color{#3D99F6}{1-\frac {x^2}{2!}+\frac {x^4}{4!}-...} - \color{#D61F06}{1-\frac {x}{1!}-\frac {x^2}{2!}-...}\right)\left(\color{#3D99F6}{1-\frac {x^2}{2!}+\frac {x^4}{4!}-...} - 1\right) - \frac {x^3}2}{x^n} \\ & = \lim_{x \to 0} \frac {\left(-x-x^2-\frac {x^3}6-...\right)\left(-\frac {x^2}{2}+\frac {x^4}{24}-\frac {x^6}{720}+...\right)-\frac {x^3}2}{x^n} \\ & = \lim_{x \to 0} \frac {\frac {x^3}2+\frac{x^4}2 + \frac {x^5}{24}-\frac{x^6}{24}+...-\frac {x^3}2}{x^n} \\ & = \lim_{x \to 0} \frac {\frac{x^4}2 + \frac {x^5}{24}-\frac{x^6}{24}+...}{x^n} \end{aligned}$

From the above, we note that $L = 0$ for $n \le 3$ , $L \to \infty$ for $n \ge 5$ and $L = \frac 12$ for $n = \boxed{4}$ .