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We are asked to evaluate the limit:

\[
\lim_{x \to 1^-} f(x)
\]

where \( f(x) \) is a piecewise function defined as:

\[
f(x) = \begin{cases}
-2x^2 - 2x & \text{if } x \leq -2 \\
3x - 2 & \text{if } -2 < x < 1 \\
x^2 + 3 & \text{if } x > 1
\end{cases}
\]

We are particularly interested in the left-hand limit as \( x \to 1^- \). In this case, \( -2 < x < 1 \), so we use the second piece of the function: \( f(x) = 3x - 2 \).

To find the limit:

\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 2) = 3(1) - 2 = 1
\]

Thus, the value of the limit is \( \boxed{1} \).

Let me know if you'd like further details or have questions. Here are five related questions:

1. How do you compute a right-hand limit for the same function?
2. What happens when \( x \to -2^+ \)?
3. How does a discontinuity affect the value of a limit?
4. How can piecewise functions have different behaviors at boundaries?
5. Can a limit exist if the left-hand and right-hand limits are different?

**Tip:** Always check which part of a piecewise function applies based on the value you're approaching.

Given the following piecewise function, evaluate lim x→1⁻ f(x).

This problem involves evaluating the limit of a piecewise function as x approaches 1 from the left. It focuses on understanding how to handle limits for different pieces of a function and applies fundamental calculus concepts.

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