Let's analyze the function $f(x) = \frac{2 + x}{x - 2}$ to determine the values of the limits at $x = 0$.

Function Analysis:

$f(x) = \frac{2 + x}{x - 2}$

(a) $\lim_{x \to 0^-} f(x)$

This is the left-hand limit as $x$ approaches 0 from the left. Since $x - 2$ is negative for $x < 2$, the function is defined in this region. We can substitute values slightly less than 0 into the function:

$\lim_{x \to 0^-} f(x) = \frac{2 + 0^-}{0^- - 2} = \frac{2}{-2} = -1$

So: $\lim_{x \to 0^-} f(x) = -1$

(b) $\lim_{x \to 0^+} f(x)$

This is the right-hand limit as $x$ approaches 0 from the right. Similarly, for $x > -2$, $x - 2$ is negative, and we substitute values slightly greater than 0:

$\lim_{x \to 0^+} f(x) = \frac{2 + 0^+}{0^+ - 2} = \frac{2}{-2} = -1$

So: $\lim_{x \to 0^+} f(x) = -1$

(c) $\lim_{x \to 0} f(x)$

The limit exists only if the left-hand and right-hand limits are equal:

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

Final Answers:

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

(b) $\lim_{x \to 0^+} f(x) = -1$

(c) $\lim_{x \to 0} f(x) = -1$

Would you like further details or have any questions?

Related Questions:

1. What is the limit $\lim_{x \to 2^-} f(x)$?
2. How do vertical asymptotes affect limits of functions?
3. Can a function have different left-hand and right-hand limits?
4. What is the significance of continuity in determining limits?
5. How do you compute limits for functions involving absolute values?

Tip: When evaluating limits, always consider the function's behavior from both the left and right sides of the point of interest.