To solve this limit problem:

$\lim_{x \to 0} \frac{e^{3x} - \sin(3x) - 1}{\cos(x) - 4x^2 - 1},$

we need to examine the behavior of the numerator and the denominator as $x \to 0$.

Step-by-Step Solution

1. Check the Limit Type (Indeterminate Form): Substitute $x = 0$ directly into the expression:

   • Numerator: $e^{3 \cdot 0} - \sin(3 \cdot 0) - 1 = 1 - 0 - 1 = 0$.
   • Denominator: $\cos(0) - 4 \cdot 0^2 - 1 = 1 - 0 - 1 = 0$.

   Since both the numerator and denominator approach $0$ as $x \to 0$, we have a $\frac{0}{0}$ indeterminate form, which suggests we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule: Differentiate the numerator and the denominator with respect to $x$.

   • Numerator's derivative: $\frac{d}{dx}(e^{3x} - \sin(3x) - 1) = 3e^{3x} - 3\cos(3x).$

   • Denominator's derivative: [ \frac{d}{