The problem you uploaded asks to find the limit:

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

Step-by-Step Solution:

1. Direct Substitution: Let's first try substituting $x = -1$ directly into the expression.

   • For the first term: $\frac{1}{x+1} = \frac{1}{-1+1} = \frac{1}{0} \quad \text{(undefined)}$

   • For the second term: $\frac{3}{x^3 + 1} = \frac{3}{(-1)^3 + 1} = \frac{3}{-1 + 1} = \frac{3}{0} \quad \text{(undefined)}$

   Both terms result in division by zero when $x = -1$, so direct substitution doesn't work. We need to apply algebraic simplifications to resolve this.

2. Factoring and Simplifying:

   • The second denominator $x^3 + 1$ can be factored using the sum of cubes formula: $x^3 + 1 = (x + 1)(x^2 - x + 1)$ Thus, the second term becomes: $\frac{3}{x^3 + 1} = \frac{3}{(x + 1)(x^2 - x + 1)}$

3. Common Denominator:

   To simplify the expression, we'll combine the two terms into a single fraction. The common denominator will be $(x + 1)(x^2 - x + 1)$: $\frac{1}{x+1} - \frac{3}{(x + 1)(x^2 - x + 1)} = \frac{(x^2 - x + 1) - 3}{(x + 1)(x^2 - x + 1)}$

4. Simplifying the Numerator:

   Now simplify the numerator: $(x^2 - x + 1) - 3 = x^2 - x - 2$ This gives us the new expression: $\frac{x^2 - x - 2}{(x + 1)(x^2 - x + 1)}$

   Notice that the numerator $x^2 - x - 2$ can be factored as: $x^2 - x - 2 = (x - 2)(x + 1)$

   Therefore, the entire expression becomes: $\frac{(x - 2)(x + 1)}{(x + 1)(x^2 - x + 1)}$

5. Canceling Terms:

   We can cancel the common factor $(x + 1)$ from the numerator and denominator: $\frac{x - 2}{x^2 - x + 1}$

6. Substituting $x = -1$:

   Now, substitute $x = -1$ into the simplified expression: $\frac{-1 - 2}{(-1)^2 - (-1) + 1} = \frac{-3}{1 + 1 + 1} = \frac{-3}{3} = -1$

Final Answer:

$\lim_{{x \to -1}} \left( \frac{1}{x+1} - \frac{3}{x^3+1} \right) = -1$

Let me know if you would like further details or have any questions!

Related Questions:

1. What is the sum of cubes formula, and how is it derived?
2. Why does direct substitution result in an undefined expression at first?
3. What is the significance of factoring in limits involving rational functions?
4. How can L'Hopital's Rule be applied to indeterminate forms like this?
5. Can this type of limit be solved using a different method, like series expansion?

Tip:

Always check for factorizations or cancellations when you encounter undefined forms in limits!