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To evaluate the limit

\[
\lim_{{x \to \infty}} \left( \ln(2x) - \ln(x+1) \right)
\]

using L'Hôpital's Rule, let's first simplify the expression.

### Step 1: Rewrite the Expression
The limit can be rewritten as:

\[
\lim_{{x \to \infty}} \ln\left(\frac{2x}{x+1}\right)
\]

### Step 2: Simplify the Fraction Inside the Logarithm
Now, we simplify \( \frac{2x}{x+1} \):

\[
\frac{2x}{x+1} = \frac{2x}{x \left(1 + \frac{1}{x}\right)} = \frac{2}{1 + \frac{1}{x}}
\]

As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so the fraction approaches:

\[
\frac{2}{1+0} = 2
\]

### Step 3: Apply the Limit
Now, the limit becomes:

\[
\lim_{{x \to \infty}} \ln\left(\frac{2x}{x+1}\right) = \ln(2)
\]

### Final Answer
\[
\lim_{{x \to \infty}} \left( \ln(2x) - \ln(x+1) \right) = \ln(2)
\]

Would you like more details on any part of this solution or have any questions?

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### Related Questions to Explore:
1. What is L'Hôpital's Rule and when can it be applied?
2. How do logarithmic properties simplify complex limits?
3. How do limits behave at infinity for rational functions?
4. Can L'Hôpital’s Rule be applied multiple times in a problem?
5. How does one evaluate limits involving indeterminate forms?

### Tip:
Remember that logarithmic properties, like \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), are often useful in simplifying complex expressions before applying calculus techniques.

Use L'Hôpital's Rule to evaluate the limit: lim (ln(2x) - ln(x + 1)) as x approaches infinity.

Learn how to evaluate the limit lim (ln(2x) - ln(x + 1)) as x approaches infinity using L'Hôpital's Rule. This solution simplifies the expression using logarithmic properties and explores indeterminate forms. Ideal for calculus students.

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