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The given piecewise function is defined as:

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

We are asked to find the values of \( a \) for which \( \lim_{x \to a} f(x) \) exists. This requires the left-hand limit and right-hand limit of \( f(x) \) to be equal at any point \( a \).

### Step-by-Step Analysis

1. **For \( a < -1 \):** 
   Since the function \( f(x) = 1 + x \) is the only expression defined for \( x < -1 \), it is continuous and the limit exists for all \( a < -1 \). Hence, \( \lim_{x \to a} f(x) = 1 + a \) exists for any \( a < -1 \).

2. **At \( a = -1 \):**
   - **Left-hand limit:** As \( x \to -1^- \), we use \( f(x) = 1 + x \), so:
     \[
     \lim_{x \to -1^-} f(x) = 1 + (-1) = 0.
     \]
   - **Right-hand limit:** As \( x \to -1^+ \), we use \( f(x) = x^2 \), so:
     \[
     \lim_{x \to -1^+} f(x) = (-1)^2 = 1.
     \]
   Since \( \lim_{x \to -1^-} f(x) = 0 \) and \( \lim_{x \to -1^+} f(x) = 1 \), the limit does not exist at \( a = -1 \).

3. **For \( -1 < a < 1 \):**
   In this interval, \( f(x) = x^2 \), which is a continuous function. Therefore, the limit exists for any \( a \) in the interval \( -1 < a < 1 \), and \( \lim_{x \to a} f(x) = a^2 \).

4. **At \( a = 1 \):**
   - **Left-hand limit:** As \( x \to 1^- \), we use \( f(x) = x^2 \), so:
     \[
     \lim_{x \to 1^-} f(x) = 1^2 = 1.
     \]
   - **Right-hand limit:** As \( x \to 1^+ \), we use \( f(x) = 2 - x \), so:
     \[
     \lim_{x \to 1^+} f(x) = 2 - 1 = 1.
     \]
   Since \( \lim_{x \to 1^-} f(x) = 1 \) and \( \lim_{x \to 1^+} f(x) = 1 \), the limit exists at \( a = 1 \), and \( \lim_{x \to 1} f(x) = 1 \).

5. **For \( a > 1 \):**
   Since \( f(x) = 2 - x \) is the only expression defined for \( x > 1 \), the function is continuous and the limit exists for all \( a > 1 \). Hence, \( \lim_{x \to a} f(x) = 2 - a \) exists for any \( a > 1 \).

### Conclusion:
The limit \( \lim_{x \to a} f(x) \) exists for all \( a \) except \( a = -1 \). Therefore, the values of \( a \) for which the limit exists are:

\[
a \in (-\infty, -1) \cup (-1, 1] \cup (1, \infty).
\]

Would you like more details on any step?

### Related Questions:
1. How is the definition of continuity used to verify the existence of limits?
2. What is the significance of one-sided limits in piecewise functions?
3. How can we extend this concept to higher-dimensional piecewise functions?
4. What role does differentiability play in the behavior of limits at the points of discontinuity?
5. Can you show an example where a limit exists but the function is not continuous at that point?

**Tip:** When evaluating limits at points where a piecewise function changes definition, always consider both the left-hand and right-hand limits.

Determine the values of a for which the limit as x approaches a of f(x) exists. The piecewise function f(x) is defined as:
1 + x if x < -1,
x^2 if -1 ≤ x < 1,
2 - x if x ≥ 1.

This problem explores the limits of a piecewise function f(x) with three cases. Learn how to determine the values of a for which the limit of f(x) exists by analyzing the left-hand and right-hand limits, applying the continuity theorem.

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