The given mathematical expression is:

$\lim_{x \to \infty} (\ln x)^{\frac{1}{x}}.$

Solution:

1. Let $y = (\ln x)^{\frac{1}{x}}$. Taking the natural logarithm on both sides: $\ln y = \frac{1}{x} \ln(\ln x).$

2. Now, consider the limit of $\ln y$: $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln(\ln x)}{x}.$

   Since $\ln(\ln x)$ grows slower than $x$ as $x \to \infty$, the numerator grows much slower than the denominator.

3. Applying the growth comparison: $\frac{\ln(\ln x)}{x} \to 0 \quad \text{as } x \to \infty.$

4. Therefore: $\ln y \to 0 \quad \text{as } x \to \infty.$

   Exponentiating back to get $y$: $y \to e^0 = 1.$

Final Answer:

$\lim_{x \to \infty} (\ln x)^{\frac{1}{x}} = 1.$

Would you like a step-by-step breakdown or any clarifications? Let me know!

Here are 5 related questions for you to explore:

1. How does the growth of $\ln x$ compare to $x$ as $x \to \infty$?
2. Why is $\ln(\ln x)$ considered a slowly growing function?
3. What are some common techniques to evaluate limits involving logarithmic functions?
4. How do you use L'Hôpital's Rule to solve similar limits with $\frac{f(x)}{g(x)}$ as $x \to \infty$?
5. What is the significance of exponential functions in analyzing limits like this?

Tip: Always take the logarithm of an expression if it simplifies the exponents, especially in limits and growth rate comparisons.