If $\lim_{x \to 0} \dfrac{\cos^2 x - \cos x - e^x \cos x + e^x - \dfrac{x^3}{2}}{x^n} = L$ (where $L$ is non zero finite), then:
Step-by-Step Solution
Key Concept: When the numerator is an indeterminate form of type 0/0 under repeated cancellation, use Taylor series expansion up to sufficient order. The trick here is algebraic factoring first — recognizing $(\cos x - 1)(\cos x - e^x)$ — which collapses the $x^3$ terms cleanly, exposing the leading $x^4$ behavior without needing to expand each term blindly to high order.
**Step 1: Factor the numerator**
Group the first four terms: $\cos x(\cos x - 1) - e^x(\cos x - 1) - \dfrac{x^3}{2} = (\cos x - 1)(\cos x - e^x) - \dfrac{x^3}{2}$
**Step 2: Taylor expand each factor**
$\cos x - 1 = -\dfrac{x^2}{2} - \dfrac{x^4}{24} - \cdots$, and $\cos x - e^x = -x - x^2 - \dfrac{x^3}{6} - \cdots$
**Step 3: Multiply the two factors**
$(\cos x - 1)(\cos x - e^x) = \left(-\dfrac{x^2}{2}\right)(-x - x^2 - \cdots) = \dfrac{x^3}{2} + \dfrac{x^4}{2} + \cdots$
**Step 4: Subtract $x^3/2$**
Numerator $= \dfrac{x^3}{2} + \dfrac{x^4}{2} + \cdots - \dfrac{x^3}{2} = \dfrac{x^4}{2} + O(x^5)$
**Step 5: Identify n and L**
For limit to be non-zero finite, $n = 4$ and $L = \dfrac{1}{2}$
Correct Answer: 1, 4