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We are given a piecewise function:

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

We need to evaluate the limit:

\[
\lim_{{x \to -3}} f(x)
\]

### Step 1: Evaluate the Left-Hand Limit \(\lim_{{x \to -3^-}} f(x)\)
For \(x < -3\), the function is defined as \(f(x) = x^2 - 1\). So, we compute:

\[
\lim_{{x \to -3^-}} f(x) = \lim_{{x \to -3^-}} (x^2 - 1)
= (-3)^2 - 1 = 9 - 1 = 8
\]

### Step 2: Evaluate the Right-Hand Limit \(\lim_{{x \to -3^+}} f(x)\)
For \(x > -3\), the function is defined as \(f(x) = 2x\). Thus, we compute:

\[
\lim_{{x \to -3^+}} f(x) = \lim_{{x \to -3^+}} 2x = 2(-3) = -6
\]

### Step 3: Conclusion
Since the left-hand limit \(\lim_{{x \to -3^-}} f(x) = 8\) and the right-hand limit \(\lim_{{x \to -3^+}} f(x) = -6\), the two limits are not equal. Therefore, the limit does not exist at \(x = -3\).

The final result is:
\[
\lim_{{x \to -3}} f(x) \text{ does not exist.}
\]

Would you like further details or explanations? Here are some related questions to expand on the topic:

1. What is the significance of one-sided limits?
2. How do piecewise functions behave at boundary points?
3. Can a limit exist if the left-hand and right-hand limits are different?
4. How can discontinuities in a function be classified?
5. What is the difference between a limit existing and a function being continuous?

**Tip:** When calculating limits for piecewise functions, always check the left-hand and right-hand limits separately to ensure continuity.

Given the function f(x) = {2x if x > -3, x^2 - 1 if x < -3}, find the limit of f(x) as x approaches -3.

Explore how to find the limit of a piecewise function as x approaches -3. This problem involves calculating one-sided limits and determining if a limit exists, crucial for understanding continuity and discontinuities in functions. Suitable for high school calculus students.

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