In the expression you provided, we are examining a limit as $x \to 0^+$ involving the function $f(x)$. Specifically, it looks like the expression is:

$\lim_{x \to 0^+} f(x) = 1 + \lim_{x \to 0^+} \frac{\ln(2x)}{\frac{1}{x}}$

This simplifies to:

$= 1 + \lim_{x \to 0^+} \frac{\ln(2x)}{1/x}$

which can be further simplified as:

$= 1 + \lim_{x \to 0^+} x \cdot \ln(2x)$

Now, let's break down what happens here:

1. As $x \to 0^+$, $\ln(2x)$ approaches $-\infty$ because $\ln(2x)$ diverges negatively as $x$ gets close to zero from the right.
2. Meanwhile, $x$ approaches $0$ from the positive side.

Thus, we have a limit that involves $0 \cdot (-\infty)$, which is an indeterminate form. To evaluate this, we can apply L'Hôpital's Rule after rewriting the expression:

$\lim_{x \to 0^+} x \ln(2x) = \lim_{x \to 0^+} \frac{\ln(2x)}{1/x}$

Applying L'Hôpital's Rule involves differentiating the numerator and the denominator:

1. The derivative of $\ln(2x)$ with respect to $x$ is $\frac{1}{2x} \cdot 2 = \frac{1}{x}$.
2. The derivative of $1/x$ with respect to $x$ is $-\frac{1}{x^2}$.

So we have:

$\lim_{x \to 0^+} \frac{\ln(2x)}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$

Thus, the entire expression becomes:

$\lim_{x \to 0^+} f(x) = 1 + 0 = 1$

So, the final answer is $1$. The $\frac{\infty}{\infty}$ form observed initially was an intermediate form that required L'Hôpital's Rule to resolve.

Would you like further clarification on any step?

Here are some related questions that might interest you:

1. How does L'Hôpital's Rule help resolve indeterminate forms?
2. What other indeterminate forms are there besides $\frac{\infty}{\infty}$?
3. Can we apply L'Hôpital’s Rule multiple times in a row?
4. What are alternative methods to solve $0 \cdot \infty$ indeterminate limits?
5. How do logarithmic transformations help in limit evaluation?

Tip: When encountering an indeterminate form, try to rewrite it into a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form to apply L'Hôpital’s Rule.