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To evaluate the limit 

\[
\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}}
\]

we'll analyze the behavior of the function as \( x \) approaches infinity.

### Step 1: Apply L'Hôpital's Rule
Since both \( \ln x \) and \( 2\sqrt{x} \) grow without bounds as \( x \to \infty \), we have an indeterminate form of \( \frac{\infty}{\infty} \), which allows us to apply L'Hôpital's Rule. This rule states that for functions of the form \( \frac{f(x)}{g(x)} \) where both \( f(x) \to \infty \) and \( g(x) \to \infty \) as \( x \to \infty \), we can find the limit by differentiating the numerator and the denominator.

### Step 2: Differentiate the Numerator and the Denominator
1. Differentiate the numerator, \( \ln x \):
   \[
   \frac{d}{dx}(\ln x) = \frac{1}{x}
   \]

2. Differentiate the denominator, \( 2\sqrt{x} \):
   \[
   \frac{d}{dx}(2\sqrt{x}) = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}
   \]

### Step 3: Rewrite the Limit Using the Derivatives
The limit now becomes:
\[
\lim_{{x \to \infty}} \frac{\frac{1}{x}}{\frac{1}{\sqrt{x}}} = \lim_{{x \to \infty}} \frac{1}{x} \cdot \sqrt{x} = \lim_{{x \to \infty}} \frac{\sqrt{x}}{x}
\]

### Step 4: Simplify the Expression
Rewrite \( \frac{\sqrt{x}}{x} \) as:
\[
\frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x} = x^{-1/2} = \frac{1}{\sqrt{x}}
\]

### Step 5: Take the Limit
Now, as \( x \to \infty \), \( \frac{1}{\sqrt{x}} \to 0 \).

Therefore:
\[
\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} = 0
\]

### Final Answer
\[
\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} = 0
\]

Would you like more details or have any questions?

Here are five related questions for further exploration:
1. What is L'Hôpital's Rule and when can it be applied?
2. How does the growth rate of \( \ln x \) compare to \( x^{\alpha} \) for various values of \( \alpha \)?
3. Can we generalize this limit if the denominator is changed to \( x^p \) for some \( p > 0 \)?
4. What are other methods to analyze limits of the form \( \frac{\ln x}{x^p} \) as \( x \to \infty \)?
5. How does the behavior of logarithmic functions compare to polynomial and exponential functions as \( x \) approaches infinity?

**Tip:** For limits involving \( \ln x \) and powers of \( x \), it can often help to compare their growth rates, as polynomial functions tend to grow faster than logarithmic functions.

Find \( \lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} \).

Explore the solution to finding the limit of \( \frac{\ln x}{2\sqrt{x}} \) as \( x \to \infty \). This problem demonstrates the application of L'Hôpital's Rule and differentiation techniques in calculus to analyze the growth behavior of logarithmic and radical functions. Suitable for undergraduate calculus students.

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