To solve the limit:

$\lim_{x \to 0} \frac{\ln(1 + 5x)}{x}$

Step-by-Step Solution:

1. Identify the indeterminate form:
   As $x \to 0$, the natural logarithm argument $1 + 5x \to 1$, so:
   $\ln(1 + 5x) \to \ln(1) = 0$ Thus, the expression becomes $\frac{0}{0}$, which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule:
   L'Hôpital's Rule states that for indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$:
   $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ if the derivative limits exist.

   Here:

   • $f(x) = \ln(1 + 5x)$
   • $g(x) = x$

   Now, differentiate $f(x)$ and $g(x)$ with respect to $x$:

   • $f'(x) = \frac{d}{dx} \ln(1 + 5x) = \frac{5}{1 + 5x}$
   • $g'(x) = \frac{d}{dx} x = 1$

3. Rewrite the limit with the derivatives:
   Using L'Hôpital's Rule:
   $\lim_{x \to 0} \frac{\ln(1 + 5x)}{x} = \lim_{x \to 0} \frac{\frac{5}{1 + 5x}}{1}$

   Simplify the expression:
   $\lim_{x \to 0} \frac{5}{1 + 5x}$

4. Evaluate the limit:
   As $x \to 0$, the denominator $1 + 5x \to 1$. Therefore:
   $\lim_{x \to 0} \frac{5}{1 + 5x} = \frac{5}{1} = 5$

Final Answer:

$\boxed{5}$

Would you like a detailed explanation of L'Hôpital's Rule or related concepts? Here are 5 related questions to expand your understanding:

1. What are the conditions for applying L'Hôpital's Rule?
2. Can we solve the same limit without using L'Hôpital's Rule?
3. How do you differentiate logarithmic functions in general?
4. What happens to $\ln(1 + kx)$ as $x \to 0$ for different constants $k$?
5. How does the behavior of logarithmic and exponential limits differ as $x \to 0$?

Tip: Always check for indeterminate forms before applying L'Hôpital's Rule!