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We are asked to find:

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

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

\[
f(x) = 
\begin{cases} 
x^2 + 2 & \text{for } x \leq -1, \\
2x - 5 & \text{for } x > -1.
\end{cases}
\]

### Steps to solve:

#### 1. **Left-hand limit (\( x \to -1^- \))**:
Since \( x \to -1^- \) approaches from the left (values of \( x \leq -1 \)), we use the function \( f(x) = x^2 + 2 \).
\[
\lim_{x \to -1^-} f(x) = (-1)^2 + 2 = 1 + 2 = 3.
\]

#### 2. **Right-hand limit (\( x \to -1^+ \))**:
For \( x \to -1^+ \), approaching from the right (values of \( x > -1 \)), we use \( f(x) = 2x - 5 \).
\[
\lim_{x \to -1^+} f(x) = 2(-1) - 5 = -2 - 5 = -7.
\]

#### 3. **Conclusion**:
The left-hand limit is 3, and the right-hand limit is -7. Since the left and right limits are not equal, the two-sided limit does not exist.

Thus, the correct answer is:
\[
\boxed{d. \text{Does not exist}}
\]

Would you like any further explanation or details? Here are 5 follow-up questions to deepen understanding:

1. What is the significance of left-hand and right-hand limits in piecewise functions?
2. Can a function be continuous if its two-sided limit does not exist?
3. How do we determine continuity at a point in a piecewise function?
4. What happens if the left-hand and right-hand limits are equal?
5. Could this piecewise function be modified to make the limit exist?

**Tip:** Always check both the left-hand and right-hand limits when working with piecewise functions!

Let f(x) = { x^2 + 2 if x ≤ -1; 2x - 5 if x > -1 }. Find lim f(x) as x approaches -1.

Learn how to find the limit of a piecewise function as x approaches -1. This problem involves calculating left-hand and right-hand limits to determine if the two-sided limit exists. Perfect for high school and early college students studying calculus.

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