Find the `Lim_(nrarroo) ((""^(3n)C_(n))/(""^(2n)C_(n)))^(1/n)`
Where `""^(i)C_(j)` is a binomial coefficient which means `(i.(i-1)"…."(i-j+1))/(j.(j-1)"….."2.1)`

Answer

To solve the limit \( \lim_{n \to \infty} \left( \frac{{{3n \choose n}}}{{{2n \choose n}}} \right)^{\frac{1}{n}} \), we can follow these steps:

### Step 1: Rewrite the Limit
We start by rewriting the limit:
\[ L = \lim_{n \to \infty} \left( \frac{{{3n \choose n}}}{{{2n \choose n}}} \right)^{\frac{1}{n}} \]

### Step 2: Take the Natural Logarithm
Taking the natural logarithm on both sides:
\[ \ln L = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{{{3n \choose n}}}{{{2n \choose n}}} \right) \]

### Step 3: Expand the Binomial Coefficients
Using the formula for binomial coefficients:
\[ {n \choose k} = \frac{n!}{k!(n-k)!} \]
we can express the binomial coefficients:
\[ {3n \choose n} = \frac{(3n)!}{n!(2n)!}, \quad {2n \choose n} = \frac{(2n)!}{n!n!} \]
Thus,
\[ \frac{{{3n \choose n}}}{{{2n \choose n}}} = \frac{(3n)! \cdot n! \cdot n!}{(2n)! \cdot n! \cdot (2n)!} = \frac{(3n)!}{(2n)! \cdot n!} \]

### Step 4: Substitute Back
Substituting back into the logarithm:
\[ \ln L = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{(3n)!}{(2n)! \cdot n!} \right) \]

### Step 5: Apply Stirling's Approximation
Using Stirling's approximation \( n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \):
\[ (3n)! \sim \sqrt{6 \pi n} \left( \frac{3n}{e} \right)^{3n}, \quad (2n)! \sim \sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}, \quad n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]

### Step 6: Substitute Stirling's Approximation
Substituting these approximations into the logarithm:
\[ \ln L = \lim_{n \to \infty} \frac{1}{n} \left( \ln \left( \sqrt{6 \pi n} \left( \frac{3n}{e} \right)^{3n} \right) - \ln \left( \sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n} \right) - \ln \left( \sqrt{2 \pi n} \left( \frac{n}{e} \right)^{n} \right) \right) \]

### Step 7: Simplify the Logarithm
Breaking it down:
\[ \ln L = \lim_{n \to \infty} \frac{1}{n} \left( \frac{1}{2} \ln(6 \pi n) + 3n \ln(3n) - 3n - \frac{1}{2} \ln(4 \pi n) - 2n \ln(2n) + 2n - \frac{1}{2} \ln(2 \pi n) - n \ln(n) + n \right) \]

### Step 8: Collect Terms
Collecting terms and simplifying:
\[ \ln L = \lim_{n \to \infty} \frac{1}{n} \left( \left(3 \ln 3 - 2 \ln 2 - \ln 2\right)n + \text{lower order terms} \right) \]
This simplifies to:
\[ \ln L = \lim_{n \to \infty} \left( (3 \ln 3 - 3 \ln 2) + \text{lower order terms} \right) \]

### Step 9: Final Calculation
Thus,
\[ L = e^{3 \ln 3 - 3 \ln 2} = \frac{27}{8} \]

### Conclusion
The final answer is:
\[ L = \frac{27}{8} \]