Compute the limit `A = underset(n rarr oo)("lim") (root(n)(n!))/(n)`

Answer

To compute the limit \( A = \lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the expression for \( A \):
\[ A = \lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} \]
This can be rewritten as:
\[ A = \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \]

### Step 2: Apply the Logarithm
To simplify the limit, we take the natural logarithm:
\[ \log A = \lim_{n \to \infty} \left( \frac{1}{n} \log(n!) - \log n \right) \]

### Step 3: Use Stirling's Approximation
Using Stirling's approximation, we know that:
\[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]
Taking the logarithm gives:
\[ \log(n!) \sim \frac{1}{2} \log(2 \pi n) + n \log n - n \]
Thus, we can write:
\[ \frac{1}{n} \log(n!) \sim \frac{1}{2n} \log(2 \pi n) + \log n - 1 \]

### Step 4: Substitute Back into the Limit
Substituting this back into our expression for \( \log A \):
\[ \log A = \lim_{n \to \infty} \left( \frac{1}{2n} \log(2 \pi n) + \log n - 1 - \log n \right) \]
This simplifies to:
\[ \log A = \lim_{n \to \infty} \left( \frac{1}{2n} \log(2 \pi n) - 1 \right) \]

### Step 5: Evaluate the Limit
As \( n \to \infty \), \( \frac{1}{2n} \log(2 \pi n) \) approaches 0. Therefore:
\[ \log A = 0 - 1 = -1 \]

### Step 6: Exponentiate to Find \( A \)
Exponentiating both sides gives:
\[ A = e^{-1} = \frac{1}{e} \]

### Final Answer
Thus, the limit is:
\[ A = \frac{1}{e} \]

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