Let's examine the image you've provided. It appears to ask for the limits of a function at specific points based on its graph.

For $x = -8$:

You are asked to find:

1. $\lim_{x \to -8^-} f(x)$ (the left-hand limit as $x$ approaches -8)
2. $\lim_{x \to -8^+} f(x)$ (the right-hand limit as $x$ approaches -8)
3. $\lim_{x \to -8} f(x)$ (the two-sided limit as $x$ approaches -8)

For $x = 1$:

You are asked to find:

1. $\lim_{x \to 1^-} f(x)$ (the left-hand limit as $x$ approaches 1)
2. $\lim_{x \to 1^+} f(x)$ (the right-hand limit as $x$ approaches 1)
3. $\lim_{x \to 1} f(x)$ (the two-sided limit as $x$ approaches 1)

Analyzing the Graph:

Let’s analyze the graph to determine these limits.

At $x = -8$:

• Left-hand limit: As $x \to -8^-$, the function seems to approach the value of 3 (based on the plotted points).
• Right-hand limit: As $x \to -8^+$, the function also seems to approach 3.
• Two-sided limit: Since both one-sided limits are equal, $\lim_{x \to -8} f(x) = 3$.

At $x = 1$:

• Left-hand limit: As $x \to 1^-$, the function approaches 0.
• Right-hand limit: As $x \to 1^+$, the function approaches a different value, which appears to be 2.
• Two-sided limit: Since the left-hand and right-hand limits are not equal, the two-sided limit does not exist.

Final Answers:

1. $\lim_{x \to -8^-} f(x) = 3$

2. $\lim_{x \to -8^+} f(x) = 3$

3. $\lim_{x \to -8} f(x) = 3$

4. $\lim_{x \to 1^-} f(x) = 0$

5. $\lim_{x \to 1^+} f(x) = 2$

6. $\lim_{x \to 1} f(x)$ does not exist.

Would you like further clarification or details? Here are some related questions:

1. What happens when the left-hand and right-hand limits differ?
2. How can discontinuities affect the existence of limits?
3. What is the relationship between limits and continuity at a point?
4. How would the graph look if the two-sided limit at $x = 1$ existed?
5. How do we interpret limits at points where the function is not defined?

Tip: To find the two-sided limit, ensure both the left-hand and right-hand limits match. If they don't, the two-sided limit does not exist.