Let's analyze the graph and the given statements one by one.

Statement I:

$\lim_{{x \to 2^-}} g(x) = \lim_{{x \to 2^+}} g(x)$

From the graph, as $x \to 2^-$ (approaching from the left) and $x \to 2^+$ (approaching from the right), both limits approach 1. Therefore, the left-hand and right-hand limits are equal: $\lim_{{x \to 2}} g(x) = 1$ Thus, Statement I is true.

Statement II:

$\lim_{{x \to 6}} g(x) \text{ does not exist}$

From the graph, as $x \to 6^-$ (from the left), $g(x) \to 2$, and as $x \to 6^+$ (from the right), $g(x) \to -2$. Since the left-hand and right-hand limits are not equal, the limit does not exist. Therefore, Statement II is true.

Statement III:

$\lim_{{x \to 2}} g(x) = g(2)$

From the graph, we see:

• $g(2) = 3$
• $\lim_{{x \to 2}} g(x) = 1$ (as calculated from Statement I).

Since the limit as $x \to 2$ is not equal to $g(2)$, Statement III is false.

Conclusion:

The true statements are I and II. Therefore, the correct answer is (C) II and III only.

Would you like more details or further explanations? Here are some additional related questions:

1. What does it mean for a limit to exist at a point?
2. How can you visually determine if a function is continuous at a point?
3. What is the significance of a jump discontinuity?
4. How do left-hand and right-hand limits work in piecewise functions?
5. Can a function have a limit at a point where it is not defined?

Tip: Always check both the left-hand and right-hand limits at a given point to ensure whether the overall limit exists.