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.