$lim_{n\rightarrow\infty}[(1+\frac{1}{n})^{\frac{1}{n}}(1+\frac{2}{n})^{\frac{1}{n}}.........(1+\frac{n}{n})^{\frac{1}{n}}]=$

Question

Accepted Answer

To solve the given limit problem, we need to evaluate the expression:

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

This expression can be rewritten as:

$$
\lim_{n \rightarrow \infty} \left[\prod_{k=1}^{n} \left(1+\frac{k}{n}\right)^{\frac{1}{n}}\right]
$$

Taking the natural logarithm of the expression inside the limit, we have:

$$
\ln \left(\prod_{k=1}^{n} \left(1+\frac{k}{n}\right)^{\frac{1}{n}}\right) = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right)
$$

As $ n \to \infty $, the sum $\frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right)$ resembles a Riemann sum for the integral:

$$
\int_{0}^{1} \ln(1+x) \, dx
$$

Calculating this integral, we have:

$$
\int \ln(1+x) \, dx = (1+x)\ln(1+x) - x + C
$$

Evaluating from 0 to 1:

$$
\left[(1+1)\ln(1+1) - 1\right] - \left[(1+0)\ln(1+0) - 0\right] = [2\ln 2 - 1] - [0] = 2\ln 2 - 1
$$

Thus, the limit of the logarithm is $2\ln 2 - 1$, and therefore the original limit is:

$$
e^{2\ln 2 - 1} = \frac{e^{2\ln 2}}{e} = \frac{4}{e}
$$

This confirms that the correct answer is indeed Option 4: $\frac{4}{e}$.

**Explanation for Other Options:**

1. **Option 1 (1):** This would imply that the product of the terms converges to 1, which is not the case as shown by the integral evaluation.

2. **Option 2 ($\frac{2}{e}$):** This would suggest that the integral evaluates to $\ln 2 - 1$, which is incorrect as the correct evaluation is $2\ln 2 - 1$.

3. **Option 3 ($\frac{3}{e}$):** This would imply an integral evaluation of $3\ln 2 - 1$, which is also incorrect.

In conclusion, the correct evaluation of the limit leads to Option 4, $\frac{4}{e}$, being the correct answer.