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.