1. The problem gives us the limits and function value at $x=2$ for the function $f(x)$ with domain $-2 \leq x < 4$. 2. We are asked to analyze the behavior of $f(x)$ near $x=2$ and understand continuity and limits. 3. The left-hand limit as $x$ approaches 2 is given as: $$\lim_{x \to 2^-} f(x) = 2$$ This means as $x$ approaches 2 from values less than 2, $f(x)$ approaches 2. 4. The right-hand limit as $x$ approaches 2 is: $$\lim_{x \to 2^+} f(x) = -2$$ This means as $x$ approaches 2 from values greater than 2, $f(x)$ approaches -2. 5. The function value at 2 is: $$f(2) = 0$$ 6. For a function to be continuous at $x=2$, the left-hand limit, right-hand limit, and function value must all be equal. 7. Here, the left and right limits are not equal ($2 \neq -2$), so the limit at $x=2$ does not exist (the two-sided limit does not exist). 8. Also, $f(2) = 0$ is different from both limits, so $f$ is not continuous at $x=2$. 9. The domain $-2 \leq x < 4$ means the function is defined from $-2$ up to but not including 4. 10. Summary: The function has a jump discontinuity at $x=2$ because the left and right limits differ and the function value is different from both. Final answer: The limit $\lim_{x \to 2} f(x)$ does not exist due to differing one-sided limits, and $f$ is not continuous at $x=2$.