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Let's evaluate both limits given in the problem.

### Part (a):  
We are asked to find:
\[
\lim_{x \to 0} \frac{x^2 - 4}{x^2 - 3x - 10}
\]

1. **Factor the numerator and denominator**:
   The numerator \(x^2 - 4\) is a difference of squares:
   \[
   x^2 - 4 = (x - 2)(x + 2)
   \]

   The denominator \(x^2 - 3x - 10\) can be factored by finding two numbers that multiply to \(-10\) and add to \(-3\):
   \[
   x^2 - 3x - 10 = (x - 5)(x + 2)
   \]

   So, the limit becomes:
   \[
   \lim_{x \to 0} \frac{(x - 2)(x + 2)}{(x - 5)(x + 2)}
   \]

2. **Cancel common factors**:
   The factor \((x + 2)\) appears in both the numerator and the denominator, so we cancel it (assuming \(x \neq -2\)):
   \[
   \lim_{x \to 0} \frac{x - 2}{x - 5}
   \]

3. **Evaluate the limit**:
   Now substitute \(x = 0\) into the simplified expression:
   \[
   \frac{0 - 2}{0 - 5} = \frac{-2}{-5} = \frac{2}{5}
   \]

Thus, the solution to part (a) is:
\[
\boxed{\frac{2}{5}}
\]

---

### Part (b):  
We are asked to find:
\[
\lim_{x \to -2} \frac{x^2 - 4}{x^2 - 3x - 10}
\]

1. **Use the factored form** (as in part (a)):
   We already factored the numerator and denominator:
   \[
   \frac{(x - 2)(x + 2)}{(x - 5)(x + 2)}
   \]

2. **Cancel common factors**:
   The factor \((x + 2)\) appears in both the numerator and denominator, so cancel it (as long as \(x \neq -2\)):
   \[
   \lim_{x \to -2} \frac{x - 2}{x - 5}
   \]

3. **Evaluate the limit**:
   Now substitute \(x = -2\) into the simplified expression:
   \[
   \frac{-2 - 2}{-2 - 5} = \frac{-4}{-7} = \frac{4}{7}
   \]

Thus, the solution to part (b) is:
\[
\boxed{\frac{4}{7}}
\]

Would you like further clarification or a deeper explanation of any of the steps?

---

Here are 5 related questions you might find useful:
1. What is the general approach for factoring quadratic expressions?
2. How do you handle limits that result in an indeterminate form like \( \frac{0}{0} \)?
3. Can you explain the concept of removable discontinuities in limits?
4. What happens if common factors cannot be canceled in a rational expression during a limit evaluation?
5. How can you apply L'Hopital's Rule to limits when direct substitution gives \( \frac{0}{0} \)?

### Tip:
Always try factoring when you encounter limits with polynomial expressions, as it can help simplify the problem significantly!

Evaluate the following limits: (a) \( \lim_{x \to 0} \frac{x^2 - 4}{x^2 - 3x - 10} \) (b) \( \lim_{x \to -2} \frac{x^2 - 4}{x^2 - 3x - 10} \)

Learn how to evaluate the limits \( \lim_{x \to 0} \frac{x^2 - 4}{x^2 - 3x - 10} \) and \( \lim_{x \to -2} \frac{x^2 - 4}{x^2 - 3x - 10} \) by factoring the rational expression and canceling common factors. Suitable for students studying calculus or pre-calculus.

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