Evaluate the following limit. Use l'Hôpital's Rule when it is convenient and applicabl \[ \lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+2 x^{3}+12 x^{2}} \]

To evaluate the limit \[ \lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+2 x^{3}+12 x^{2}}, \] we first observe the behavior of the numerator and denominator as \(x\) approaches 0. 1. **Numerator Analysis**: - \(e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\) (using Taylor series expansion). - \(\sin x \approx x - \frac{x^3}{6} + O(x^5)\). Thus, \[ e^x - \sin x - 1 \approx \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right) - \left(x - \frac{x^3}{6} + O(x^5)\right) - 1. \] Simplifying this, we get: \[ e^x - \sin x - 1 \approx x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^3}{6} - x + O(x^4) = \frac{x^2}{2} + \frac{x^3}{3} + O(x^4). \] For very small \(x\), the dominant term in the numerator will be \(\frac{x^2}{2}\). 2. **Denominator Analysis**: - The denominator \(x^4 + 2x^3 + 12x^2\) has \(12x^2\) as the dominant term when \(x \to 0\). Thus for small \(x\), \[ x^4 + 2x^3 + 12x^2 \approx 12x^2. \] 3. **Putting Together**: Now we can rewrite the limit: \[ \lim _{x \rightarrow 0} \frac{e^{x}-\sin x-1}{x^{4}+2 x^{3}+12 x^{2}} \approx \lim _{x \rightarrow 0} \frac{\frac{x^2}{2}}{12x^2} = \lim _{x \rightarrow 0} \frac{1/2}{12} = \frac{1}{24}. \] Thus, the limit evaluates to: \[ \boxed{\frac{1}{24}}. \]