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.