To solve the given limit, we need to examine the expression carefully:

$\lim_{{x \to -1}} \left( \frac{\frac{x^3}{6} + \frac{x^2}{2} + \frac{1}{2}x + \frac{1}{6}}{\ln(-x) + e^{x+1} - 1} \right)$

Step 1: Substitute $x = -1$ to Check for Indeterminate Form

If we substitute $x = -1$ directly, we observe:

Numerator:

$\frac{(-1)^3}{6} + \frac{(-1)^2}{2} + \frac{1}{2}(-1) + \frac{1}{6} = -\frac{1}{6} + \frac{1}{2} - \frac{1}{2} + \frac{1}{6} = 0$

Denominator:

$\ln(-(-1)) + e^{-1 + 1} - 1 = \ln(1) + e^0 - 1 = 0 + 1 - 1 = 0$

Thus, the expression is of the indeterminate form $\frac{0}{0}$, which means we can apply L'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator Separately

Differentiating the Numerator:

The numerator is: $f(x) = \frac{x^3}{6} + \frac{x^2}{2} + \frac{1}{2}x + \frac{1}{6}$

Differentiating term-by-term: $f'(x) = \frac{3x^2}{6} + \frac{2x}{2} + \frac{1}{2} = \frac{x^2}{2} + x + \frac{1}{2}$

Differentiating the Denominator:

The denominator is: $g(x) = \ln(-x) + e^{x+1} - 1$

Differentiating term-by-term: $g'(x) = \frac{-1}{x} + e^{x+1} = -\frac{1}{x} + e^{x+1}$

Step 3: Apply L'Hôpital's Rule

The limit now becomes: $\lim_{{x \to -1}} \frac{\frac{x^2}{2} + x + \frac{1}{2}}{-\frac{1}{x} + e^{x+1}}$

Step 4: Substitute $x = -1$ Again

Numerator:

$\frac{(-1)^2}{2} + (-1) + \frac{1}{2} = \frac{1}{2} - 1 + \frac{1}{2} = 0$

Denominator:

$-\frac{1}{-1} + e^0 = 1 + 1 = 2$

Thus, the limit simplifies to: $\frac{0}{2} = 0$

Conclusion:

The answer is $\boxed{0}$, which corresponds to option (d): "None of the other options."

Would you like further details or have any questions?

Here are some related questions to expand on this topic:

1. What is the significance of L'Hôpital's Rule in evaluating limits?
2. How do we handle limits that involve logarithmic or exponential functions?
3. What other techniques can be used to solve limits if L'Hôpital's Rule is not applicable?
4. How does indeterminate form influence the choice of method for limit evaluation?
5. Can all indeterminate forms be resolved using L'Hôpital's Rule?

Tip: Always check for simplification opportunities before applying advanced techniques like L'Hôpital's Rule, as sometimes basic algebra can simplify the process.