We want to find the limit:

\[
\lim_{n \to \infty} \left( \binom{2n}{n} \right)^{1/n}
\]

### Step-by-Step Solution

1. **Stirling’s Approximation:**
   For large \(n\), we use Stirling's approximation for factorials:
   \[
   n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n
   \]

2. **Using Stirling's Approximation:**
   We know that:
   \[
   \binom{2n}{n} = \frac{(2n)!}{(n!)^2}
   \]
   Applying Stirling’s approximation to \( (2n)! \) and \( n! \), we get:
   \[
   (2n)! \sim \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}
   \]
   and
   \[
   n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n
   \]
   Therefore,
   \[
   \binom{2n}{n} \sim \frac{\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}}{\left(\sqrt{2\pi n} \left(\frac{n}{e}\right)^n\right)^2}
   \]

3. **Simplifying the Expression:**
   Simplifying the binomial coefficient, we get:
   \[
   \binom{2n}{n} \sim \frac{\sqrt{4\pi n}}{2\pi n} \cdot 4^n
   \]
   Which simplifies to:
   \[
   \binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}
   \]

4. **Taking the nth Root:**
   Now, we take the \(n\)-th root of this expression:
   \[
   \left(\binom{2n}{n}\right)^{1/n} \sim \left( \frac{4^n}{\sqrt{\pi n}} \right)^{1/n}
   \]
   This simplifies to:
   \[
   \left(\binom{2n}{n}\right)^{1/n} \sim 4 \left( \frac{1}{\sqrt{\pi n}} \right)^{1/n}
   \]
   As \(n \to \infty\), \( \left(\frac{1}{\sqrt{\pi n}}\right)^{1/n} \to 1\).

5. **Final Answer:**
   Thus, we conclude:
   \[
   \lim_{n \to \infty} \left( \binom{2n}{n} \right)^{1/n} = 4
   \]

### Conclusion:
\[
\lim_{n \to \infty} \left( \binom{2n}{n} \right)^{1/n} = 4
\]

Do you want further details or have any questions?

Here are 5 related questions to explore further:
1. What is Stirling’s approximation, and how does it work for factorials?
2. How is the binomial coefficient \(\binom{2n}{n}\) interpreted combinatorially?
3. Can Stirling's approximation be used to estimate smaller factorial values?
4. What are the asymptotic behaviors of binomial coefficients for large \(n\)?
5. How does this result compare with the behavior of other binomial coefficient limits?

**Tip:** Stirling's approximation is a powerful tool for approximating large factorials, commonly used in combinatorics and probability theory.

Explore the solution to the limit problem lim n → ∞ (2n choose n)^(1/n), using Stirling's approximation and combinatorics. Understand how asymptotic analysis helps simplify binomial coefficients for large values of n. Suitable for advanced mathematics students.

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