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The expression provided is:

\[
\lim_{x \to 1} \left[(x^2 - 4)(3x + 5)\right]
\]

### Step-by-Step Solution:

1. **Substitute \( x = 1 \) into the expression**:

   \[
   \lim_{x \to 1} \left[(x^2 - 4)(3x + 5)\right] = \left[(1^2 - 4)(3(1) + 5)\right]
   \]

2. **Simplify each part**:
   - \( 1^2 - 4 = 1 - 4 = -3 \)
   - \( 3(1) + 5 = 3 + 5 = 8 \)

3. **Multiply the results**:

   \[
   (-3)(8) = -24
   \]

### Final Answer:
\[
\lim_{x \to 1} \left[(x^2 - 4)(3x + 5)\right] = -24
\]

Would you like more details on any part of this process, or do you have any other questions?

Here are some related questions:
1. What is the concept of limits in calculus?
2. How does substitution work in evaluating limits?
3. What are the conditions under which a limit does not exist?
4. Can you explain the factorization of quadratic expressions?
5. How do you handle limits involving indeterminate forms like \( \frac{0}{0} \)?

**Tip:** When solving limits, always check if direct substitution is possible before applying more advanced techniques like L'Hôpital's Rule.

Learn how to evaluate the limit lim(x -> 1) [(x^2 - 4)(3x + 5)] with a step-by-step solution. This problem involves limits and substitution, covering key algebraic concepts like quadratic equations. Suitable for advanced high school students.

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