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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.

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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!

Find the limit: lim_{h→0} [f(9+h) - f(9)] / h, where f(x) = -9x - 1.

This limit problem focuses on finding the derivative of a linear function f(x) = -9x - 1 using the difference quotient formula. Learn step-by-step how to apply the limit definition of a derivative to solve this calculus problem.

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