The problem you're asking about is:

$\lim_{{x \to 3}} \frac{x^3 - 18x^2 + 81}{(x - 3)^2}$

To solve this limit, we begin by analyzing if direct substitution leads to a determinate or indeterminate form.

Step 1: Substituting $x = 3$

If we substitute $x = 3$ directly into the expression:

For the numerator: $x^3 - 18x^2 + 81 = 3^3 - 18(3^2) + 81 = 27 - 162 + 81 = -54 + 81 = 27$ For the denominator: $(x - 3)^2 = (3 - 3)^2 = 0^2 = 0$

We get $\frac{27}{0}$, which tends to infinity because the numerator is non-zero, and the denominator approaches zero. Therefore, the limit does not exist in the typical sense but diverges to infinity.

Hence, $\lim_{{x \to 3}} \frac{x^3 - 18x^2 + 81}{(x - 3)^2} = \infty$

Let me know if you'd like any clarifications!

Expanding Questions:

1. What happens if the limit approaches from the left or right side?
2. Can we factor the numerator further to simplify?
3. What is the difference between approaching infinity and limits at infinity?
4. How does one determine whether a function tends to positive or negative infinity?
5. What role does L'Hopital's Rule play in indeterminate forms?

Tip: Always check if you get an indeterminate form like $\frac{0}{0}$ when evaluating limits; this often signals the need for simplification or other methods like L'Hopital’s Rule.