The given problem asks to find the limit of the expression:

$\lim_{h \to 0} \frac{(h - 9)^2 - 81}{h}$

Let's simplify this step by step:

1. Expand the square: The numerator $(h - 9)^2$ expands as: $(h - 9)^2 = h^2 - 18h + 81$

   Now, subtract 81 from this expression: $(h - 9)^2 - 81 = h^2 - 18h + 81 - 81 = h^2 - 18h$

2. Substitute the simplified numerator: The expression now becomes: $\lim_{h \to 0} \frac{h^2 - 18h}{h}$

3. Factor out $h$ in the numerator: $\lim_{h \to 0} \frac{h(h - 18)}{h}$

4. Cancel $h$ in the numerator and denominator: $\lim_{h \to 0} (h - 18)$

5. Take the limit as $h \to 0$: $\lim_{h \to 0} (h - 18) = -18$

Thus, the limit is $-18$.

Would you like further clarification or details?

Here are 5 related questions:

1. How do you handle limits that result in indeterminate forms like $\frac{0}{0}$?
2. What is L'Hôpital's Rule and when should it be applied?
3. How would this limit change if there were an additional term in the denominator?
4. What is the difference between one-sided and two-sided limits?
5. How can we use factoring to simplify complex limit problems?

Tip: Always check for opportunities to factor expressions before applying limit laws to simplify computations!