The problem in the image is asking for the following limit:

$\lim_{{x \to -1}} \frac{x^4 - 2x - 3}{x + 1}$

Step-by-Step Solution:

1. Direct Substitution: First, substitute $x = -1$ directly into the function to check if there is any indeterminate form:

   $\frac{(-1)^4 - 2(-1) - 3}{-1 + 1} = \frac{1 + 2 - 3}{0} = \frac{0}{0}$

   Since the result is $\frac{0}{0}$, this is an indeterminate form, which means we need to simplify the expression further using algebra.

2. Factor the numerator: Let's factor the polynomial in the numerator, $x^4 - 2x - 3$. We need to look for factors of this expression. We can use synthetic division or trial-and-error with possible roots.

   Trying $x = -1$ as a root, we perform synthetic division on $x^4 - 2x - 3$ by $x + 1$. The quotient will give us the simplified form of the expression.

   After synthetic division, we get:

   $x^4 - 2x - 3 = (x + 1)(x^3 - x^2 + x - 3)$

3. Cancel the common factor: Now that we have factored the numerator, we can cancel the common factor $x + 1$ from the numerator and denominator:

   $\frac{(x + 1)(x^3 - x^2 + x - 3)}{x + 1} = x^3 - x^2 + x - 3 \quad \text{for} \quad x \neq -1$

4. Substitute $x = -1$ into the simplified expression: Now substitute $x = -1$ into the simplified expression $x^3 - x^2 + x - 3$:

   $(-1)^3 - (-1)^2 + (-1) - 3 = -1 - 1 - 1 - 3 = -6$

Thus, the value of the limit is:

$\lim_{{x \to -1}} \frac{x^4 - 2x - 3}{x + 1} = -6$

Would you like further details on any of the steps or have any questions?

Relative Questions:

1. How do you perform synthetic division to factor polynomials?
2. Why does the indeterminate form $\frac{0}{0}$ occur in limits?
3. What is the significance of canceling common factors in limits?
4. How can L'Hopital's Rule be applied to solve limits involving indeterminate forms?
5. What other methods can be used to solve limits when factoring is difficult?

Tip:

Always check for indeterminate forms by substituting the limit value first before trying to simplify the expression.