The problem asks to evaluate the limit:

$\lim_{x \to -1} \frac{x+1}{x^3+1}.$

Step-by-step solution:

1. Factorize the denominator: The denominator $x^3 + 1$ is a sum of cubes, which can be factored using the formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2).$ Here, $x^3 + 1 = (x + 1)(x^2 - x + 1)$.

2. Rewrite the limit: Substitute the factorized form into the limit: $\lim_{x \to -1} \frac{x+1}{x^3+1} = \lim_{x \to -1} \frac{x+1}{(x+1)(x^2 - x + 1)}.$

3. Simplify the fraction: Cancel the common factor $x+1$ in the numerator and denominator (provided $x \neq -1$): $\lim_{x \to -1} \frac{x+1}{(x+1)(x^2 - x + 1)} = \lim_{x \to -1} \frac{1}{x^2 - x + 1}.$

4. Substitute $x = -1$: After simplifying, substitute $x = -1$ into the denominator: $x^2 - x + 1 = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3.$ Thus, the limit becomes: $\frac{1}{3}.$

Final Answer:

The value of the limit is: $\boxed{B. \frac{1}{3}}$

Would you like further explanation or additional practice problems?

Here are 5 related questions to try:

1. Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x-2}$.
2. Factorize and simplify $\frac{x^3 - 1}{x - 1}$ as $x \to 1$.
3. Compute $\lim_{x \to 0} \frac{\sin(x)}{x}$.
4. Solve $\lim_{x \to -1} \frac{x^2 - 1}{x+1}$.
5. Determine $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$.

Tip:

When dealing with limits, always try to simplify the expression, especially by factoring, to handle indeterminate forms like $\frac{0}{0}$.