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The problem is asking to find the value of the following limit:

\[
\lim_{{x \to -1}} \left( x^2 - 2x - 3 \right)
\]

To solve this, we can substitute \(x = -1\) directly into the function \(f(x) = x^2 - 2x - 3\), since this is a polynomial and continuous at all points.

Substituting \(x = -1\):

\[
f(-1) = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0
\]

Thus, the value of the limit is:

\[
\boxed{0}
\]

Would you like further details or have any questions about this problem? Here are some related questions for further practice:

1. How do you solve limits involving rational functions with discontinuities?
2. What is the difference between evaluating a limit and solving for a function's value at a point?
3. Can limits always be solved by direct substitution?
4. How do you handle limits when the result is an indeterminate form like \(\frac{0}{0}\)?
5. What methods can be used to solve limits approaching infinity?

**Tip:** When dealing with polynomials, limits at a point can often be solved simply by substituting the value directly.

Determina cuál de las siguientes opciones representa el valor del siguiente límite: lim_{x→-1}(x^2 - 2x - 3)

Learn how to solve the limit problem lim_{x→-1}(x^2 - 2x - 3) using direct substitution. This problem covers key calculus concepts such as limits of polynomials and direct substitution, suitable for high school students.

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