Question:

Step 1: Factor the numerator. The numerator is given by: $$N(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}$$ Group terms to factor: $$N(x) = \cos x (\cos x - 1) - e^x (\cos x - 1) - \frac{x^3}{2}$$ $$N(x) = (\cos x - e^x)(\cos x - 1) - \frac{x^3}{2}$$ Step 2: Expand using Taylor series around $x=0$. The Taylor series expansions for $\cos x$ and $e^x$ around $x=0$ are: $$\begin{aligned} \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6) \\ &= 1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6) \end{aligned}$$ $$\begin{aligned} e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + O(x^5) \\ &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + O(x^5) \end{aligned}$$ Step 3: Compute the factors $\cos x - e^x$ and $\cos x - 1$. Subtracting the series for $e^x$ from $\cos x$: $$\begin{aligned} \cos x - e^x &= \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6)\right) - \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + O(x^5)\right) \\ &= -x - \left(\frac{x^2}{2} + \frac{x^2}{2}\right) - \frac{x^3}{6} + \left(\frac{x^4}{24} - \frac{x^4}{24}\right) + O(x^5) \\ &= -x - x^2 - \frac{x^3}{6} + O(x^5) \end{aligned}$$ Subtracting $1$ from $\cos x$: $$\begin{aligned} \cos x - 1 &= \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6)\right) - 1 \\ &= -\frac{x^2}{2} + \frac{x^4}{24} - O(x^6) \end{aligned}$$ Step 4: Compute the product $(\cos x - e^x)(\cos x - 1)$. Multiply the series obtained in Step 3, keeping terms up to $x^4$: $$\begin{aligned} (\cos x - e^x)(\cos x - 1) &= \left(-x - x^2 - \frac{x^3}{6} + O(x^5)\right) \left(-\frac{x^2}{2} + \frac{x^4}{24} - O(x^6)\right) \end{aligned}$$ The terms contributing to $x^3$ and $x^4$ are: $$\begin{aligned} (-x)\left(-\frac{x^2}{2}\right) &= \frac{x^3}{2} \\ (-x^2)\left(-\frac{x^2}{2}\right) &= \frac{x^4}{2} \end{aligned}$$ Thus, the product is: $$(\cos x - e^x)(\cos x - 1) = \frac{x^3}{2} + \frac{x^4}{2} + O(x^5)$$ Step 5: Substitute the product back into the numerator. Substitute the result from Step 4 into the expression for $N(x)$ from Step 1: $$\begin{aligned} N(x) &= \left(\frac{x^3}{2} + \frac{x^4}{2} + O(x^5)\right) - \frac{x^3}{2} \\ &= \frac{x^4}{2} + O(x^5) \end{aligned}$$ Step 6: Determine $n$ and $L$. The given limit is $\lim_{x \to 0} \dfrac{N(x)}{x^n} = L$. Substitute the expansion for $N(x)$: $$\lim_{x \to 0} \dfrac{\frac{x^4}{2} + O(x^5)}{x^n} = L$$ For $L$ to be a non-zero finite value, the lowest power of $x$ in the numerator must match the power in the denominator. Therefore, $n=4$. Substituting $n=4$ into the limit expression: $$\begin{aligned} L &= \lim_{x \to 0} \dfrac{\frac{x^4}{2} + O(x^5)}{x^4} \\ &= \lim_{x \to 0} \left(\frac{1}{2} + O(x)\right) \\ &= \frac{1}{2} \end{aligned}$$ Thus, $n=4$ and $L=\frac{1}{2}$, which corresponds to options 3 and 4. Therefore: $\boxed{1, 4}$