To solve this, let's complete the application of L'Hôpital's Rule to find the missing expression.

Step-by-Step Solution:

We are given: $\lim_{x \to \infty} \frac{\ln\left(1 + \frac{9}{x}\right)}{\frac{1}{5x}}$ After applying L'Hôpital's Rule once, we need to differentiate the numerator and the denominator with respect to $x$.

1. Differentiate the Numerator:

The numerator is: $\ln\left(1 + \frac{9}{x}\right)$ Using the chain rule, the derivative is: $\frac{1}{1 + \frac{9}{x}} \cdot \left(-\frac{9}{x^2}\right) = \frac{-\frac{9}{x^2}}{1 + \frac{9}{x}}$

2. Differentiate the Denominator:

The denominator is: $\frac{1}{5x}$ Differentiating with respect to $x$ gives: $-\frac{1}{5x^2}$

3. Substitute into the Limit:

Now the limit becomes: $\lim_{x \to \infty} \frac{\frac{-\frac{9}{x^2}}{1 + \frac{9}{x}}}{-\frac{1}{5x^2}}$ This simplifies to: $\lim_{x \to \infty} \frac{9}{1 + \frac{9}{x}} \cdot 5 = \lim_{x \to \infty} \frac{45}{1 + \frac{9}{x}}$ As $x \to \infty$, $\frac{9}{x} \to 0$, so we get: $\frac{45}{1} = 45$

Final Answer:

$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{9}{x}\right)}{\frac{1}{5x}} = 45$

Let me know if you'd like further details, or have any questions!

Follow-Up Questions

1. What is L'Hôpital's Rule and why is it used here?
2. How would the solution change if the $9$ in $\frac{9}{x}$ were replaced by another constant?
3. How do we interpret the result in terms of the original function $y$?
4. Can you apply L'Hôpital's Rule more than once in cases where it’s necessary?
5. How do you evaluate limits involving logarithmic functions as $x \to \infty$?

Tip

When using L'Hôpital's Rule, ensure that the limit is indeed an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying it.