https://file.aiquickdraw.com/mathbot/file/17260085630345s6t6ijq.jpg

Let's analyze the given graph to determine the following limits:

### a) \(\lim_{{x \to -2^-}} f(x)\)
This is asking for the limit of \(f(x)\) as \(x\) approaches \(-2\) from the left-hand side (i.e., \(x \to -2^-\)).

From the graph:
- As \(x\) approaches \(-2\) from the left, the function approaches the value \(1\).

Thus, \(\lim_{{x \to -2^-}} f(x) = 1\).

### b) \(\lim_{{x \to -2^+}} f(x)\)
This is asking for the limit of \(f(x)\) as \(x\) approaches \(-2\) from the right-hand side (i.e., \(x \to -2^+\)).

From the graph:
- As \(x\) approaches \(-2\) from the right, the function approaches the value \(3\).

Thus, \(\lim_{{x \to -2^+}} f(x) = 3\).

### c) \(\lim_{{x \to -2}} f(x)\)
This is asking for the two-sided limit of \(f(x)\) as \(x\) approaches \(-2\) from both sides.

Since the left-hand limit \(\lim_{{x \to -2^-}} f(x) = 1\) and the right-hand limit \(\lim_{{x \to -2^+}} f(x) = 3\), the two-sided limit **does not exist** because the left and right limits are not equal.

Thus, \(\lim_{{x \to -2}} f(x)\) **does not exist**.

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Would you like any more details or clarifications?

### Related Questions:
1. What is a one-sided limit, and how does it differ from a two-sided limit?
2. How do discontinuities in a graph affect limits?
3. What happens if the left-hand and right-hand limits are equal at a point?
4. Can limits exist at points where the function is undefined?
5. How can limits help in finding the behavior of a function near discontinuities?

### Tip:
When evaluating limits from a graph, always check the direction from which \(x\) approaches and whether the function has any jumps or breaks at that point.

Use the graph to determine the following limits: 
 a) lim(x → -2⁻) f(x) 
 b) lim(x → -2⁺) f(x) 
 c) lim(x → -2) f(x)

Learn how to evaluate one-sided and two-sided limits using a graph. This problem involves limits at x = -2, focusing on left-hand and right-hand limits, and determining when the two-sided limit does not exist. Suitable for Grades 11-12.

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