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The problem presented involves finding the limit:

\[
\lim_{{h \to 0}} \frac{e^{9h} - 1}{h}
\]

This is a standard limit in calculus, often related to the derivative of the exponential function. To solve this, let's break it down:

### Step-by-Step Solution:

1. **Recognize the form:**
   The expression \(\frac{e^{9h} - 1}{h}\) resembles the derivative of \(e^{9h}\) at \(h = 0\). Recall that the derivative of \(e^{kx}\) with respect to \(x\) is \(ke^{kx}\). 

2. **Expand using a series (if needed):**
   As \(h \to 0\), you can expand \(e^{9h}\) using the first few terms of the Taylor series expansion around \(h = 0\):
   \[
   e^{9h} = 1 + 9h + \frac{(9h)^2}{2!} + \cdots
   \]
   Subtracting 1 gives:
   \[
   e^{9h} - 1 = 9h + \frac{(9h)^2}{2!} + \cdots
   \]
   Therefore, for small \(h\):
   \[
   \frac{e^{9h} - 1}{h} = 9 + \frac{9^2 h}{2!} + \cdots
   \]
   As \(h \to 0\), higher-order terms disappear, leaving:
   \[
   \frac{e^{9h} - 1}{h} \to 9
   \]

Thus, the value of the limit is:

\[
\boxed{9}
\]

This limit evaluates to 9, which is the nearest integer as required.

---

Would you like more details on any step or have additional questions?

Here are 5 related questions for further practice:

1. What is the derivative of \(e^{kx}\) with respect to \(x\) in general?
2. How does the Taylor expansion help in approximating limits for small values of \(h\)?
3. What is the general formula for finding the limit \(\lim_{h \to 0} \frac{e^{ah} - 1}{h}\)?
4. How would the answer change if the exponent was \(5h\) instead of \(9h\)?
5. What are some other common limits that appear in calculus?

**Tip:** When dealing with exponential limits as \(h \to 0\), recognizing the derivative of exponential functions can often simplify the process.

Enter the answer to the nearest integer. lim (h -> 0) (e^(9h) - 1)/h

This calculus problem involves finding the limit of (e^(9h) - 1)/h as h approaches 0. The solution uses the derivative of exponential functions and the Taylor series expansion to simplify the limit, arriving at a final value of 9.

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