We are tasked with finding the following limit:

$\lim_{h \to 0} \frac{f(9+h) - f(9)}{h}$ where $f(x) = -9x - 1$.

Step-by-step Solution:

1. Substitute $f(x)$ into the expression:
   We know that $f(x) = -9x - 1$. So, calculate $f(9+h)$ and $f(9)$.

   $f(9+h) = -9(9+h) - 1 = -81 - 9h - 1 = -82 - 9h$

   $f(9) = -9(9) - 1 = -81 - 1 = -82$

2. Set up the difference:

   $f(9+h) - f(9) = (-82 - 9h) - (-82) = -9h$

3. Form the difference quotient:

   $\frac{f(9+h) - f(9)}{h} = \frac{-9h}{h} = -9$

4. Take the limit as $h \to 0$:

   The expression simplifies to $-9$, which is independent of $h$, so the limit is:

   $\lim_{h \to 0} \frac{f(9+h) - f(9)}{h} = -9$

Final Answer:

$\boxed{-9}$

Let me know if you need further details or clarification.

Here are 5 related questions you can explore:

1. How is the derivative of a linear function different from that of a quadratic function?
2. Can you apply the difference quotient formula to a quadratic function, and how does it change the process?
3. What is the geometric interpretation of finding the limit in this problem?
4. How would the result differ if $f(x)$ were a quadratic function instead of linear?
5. How does this problem relate to finding the slope of a tangent line at a specific point?

Tip: The expression you calculated here is essentially the derivative of the function $f(x)$ at $x = 9$, which is a foundational concept in calculus!