`lim_(n-gtoo)[(1+1/n)(1+2/n)(1+n/n)]^(1/n)`

Options

A. `4//e`
B. `e//4`
C. 4e
D. none of these

Answer

To solve the limit problem

\[ \lim_{n \to \infty} \left( (1 + \frac{1}{n})(1 + \frac{2}{n}) \cdots (1 + \frac{n}{n}) \right)^{\frac{1}{n}}, \]

we will follow these steps:

### Step 1: Define the Limit
Let
\[ L = \lim_{n \to \infty} \left( (1 + \frac{1}{n})(1 + \frac{2}{n}) \cdots (1 + \frac{n}{n}) \right)^{\frac{1}{n}}. \]

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides, we have:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \log \left( (1 + \frac{1}{n})(1 + \frac{2}{n}) \cdots (1 + \frac{n}{n}) \right). \]

Using the property of logarithms, we can separate the logarithm of the product:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \log(1 + \frac{1}{n}) + \log(1 + \frac{2}{n}) + \cdots + \log(1 + \frac{n}{n}) \right). \]

### Step 3: Rewrite as a Summation
This can be rewritten as:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \log(1 + \frac{k}{n}). \]

### Step 4: Recognize the Riemann Sum
As \( n \to \infty \), the expression becomes a Riemann sum for the integral of \( \log(1 + x) \) from 0 to 1:
\[ \log L = \int_0^1 \log(1 + x) \, dx. \]

### Step 5: Compute the Integral
To compute the integral, we can use integration by parts. Let:
- \( u = \log(1 + x) \) and \( dv = dx \).
Then, we have:
- \( du = \frac{1}{1 + x} dx \) and \( v = x \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du, \]
we get:
\[ \int_0^1 \log(1 + x) \, dx = \left[ x \log(1 + x) \right]_0^1 - \int_0^1 \frac{x}{1 + x} \, dx. \]

Evaluating the first term:
\[ \left[ x \log(1 + x) \right]_0^1 = 1 \cdot \log(2) - 0 = \log(2). \]

### Step 6: Compute the Remaining Integral
Now we compute:
\[ \int_0^1 \frac{x}{1 + x} \, dx = \int_0^1 \left( 1 - \frac{1}{1 + x} \right) dx = \left[ x - \log(1 + x) \right]_0^1 = (1 - \log(2)) - (0 - 0) = 1 - \log(2). \]

### Step 7: Combine Results
Substituting back, we have:
\[ \log L = \log(2) - (1 - \log(2)) = 2 \log(2) - 1. \]

### Step 8: Exponentiate to Find L
Thus, we find:
\[ L = e^{\log L} = e^{2 \log(2) - 1} = \frac{4}{e}. \]

### Final Answer
Therefore, the final result is:
\[ \lim_{n \to \infty} \left( (1 + \frac{1}{n})(1 + \frac{2}{n}) \cdots (1 + \frac{n}{n}) \right)^{\frac{1}{n}} = \frac{4}{e}. \]