The given mathematical expression is:

$\lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}$

Step-by-Step Solution:

1. Substitute $x = 3$ directly into the expression:

   $\frac{(3)^4 - 81}{2(3)^2 - 5(3) - 3}$

   • Numerator: $3^4 = 81$, so $81 - 81 = 0$.
   • Denominator: $2(3^2) - 5(3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0$.

   This results in a $\frac{0}{0}$ indeterminate form, which means we need to simplify the expression.

2. Factor both the numerator and the denominator:

   • The numerator $x^4 - 81$ is a difference of squares: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$
   • The denominator $2x^2 - 5x - 3$ can be factored as: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$

3. Simplify the expression by canceling out the common factor $(x - 3)$:

   $\frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)} = \frac{(x + 3)(x^2 + 9)}{2x + 1}$

4. Substitute $x = 3$ into the simplified expression:

   $\frac{(3 + 3)((3)^2 + 9)}{2(3) + 1} = \frac{6(9 + 9)}{6 + 1} = \frac{6(18)}{7} = \frac{108}{7}$

   Thus, the value of the limit is:

   $\boxed{\frac{108}{7}}$

Would you like more details or have any questions?

Related Questions:

1. What is the general approach for solving limits that result in $\frac{0}{0}$ indeterminate forms?
2. How do you factor higher-degree polynomials efficiently?
3. What is L'Hopital's Rule and when can it be applied?
4. Can all limits involving polynomials be solved by factoring?
5. How can we handle limits when direct substitution results in infinity?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ when computing limits, as these suggest the need for further simplifications or techniques like L'Hopital's Rule.