The problem provided is the following:

$f(x) = \frac{x^2 + 3x - 10}{x^2 - 4} = \frac{(x + 5)(x - 2)}{(x - 2)(x + 2)}$ You are asked to evaluate the limit:

$\lim_{x \to -2} f(x)$

Step-by-Step Solution:

1. Factor the numerator and the denominator: $f(x) = \frac{(x + 5)(x - 2)}{(x - 2)(x + 2)}$

2. Cancel the common factor $(x - 2)$: For $x \neq 2$, we can simplify the function: $f(x) = \frac{x + 5}{x + 2}$

3. Evaluate the limit: Now, we compute the limit as $x$ approaches $-2$: $\lim_{x \to -2} \frac{x + 5}{x + 2}$

4. Substitute $x = -2$ directly into the simplified expression: $\lim_{x \to -2} \frac{x + 5}{x + 2} = \frac{-2 + 5}{-2 + 2} = \frac{3}{0}$

5. Conclusion: Since dividing by zero is undefined, the limit does not exist. However, because we get a positive number divided by zero, we need to analyze the behavior as $x$ approaches $-2$ from both sides:

   • As $x \to -2^+$ (from the right), $x + 2 > 0$, leading to $+\infty$.
   • As $x \to -2^-$ (from the left), $x + 2 < 0$, leading to $-\infty$.

   Thus, the limit as $x$ approaches $-2$ is undefined due to a vertical asymptote, but we observe different behaviors from either side.

Final Answer:

$\lim_{x \to -2} f(x) = \text{undefined} \quad \text{(positive infinity from the right, negative infinity from the left)}.$

Would you like more details or have any questions?

Here are 5 related questions you might consider:

1. What happens when you evaluate the limit at $x = 2$?
2. How can you determine if a function has a vertical asymptote at a given point?
3. What is the behavior of $f(x)$ as $x \to \infty$?
4. How would the answer change if you approached a different type of indeterminate form (e.g., $0/0$)?
5. Can this function be redefined to remove the discontinuity at $x = 2$?

Tip: When evaluating limits that result in $\frac{c}{0}$, analyze the behavior of the function from both sides to determine if there is a vertical asymptote or if the limit does not exist.