$$\lim_{n \to \infty} \frac{((n+1)(n+2) \cdots (n+n))^{1/n}}{n}.$$​

Correct option is B

​$$\text{Let } L = \lim_{n \to \infty} \frac{((n+1)(n+2) \cdots (n+n))^{1/n}}{n}. \\[10pt]\text{We can rewrite the expression as:} \\[10pt]L = \lim_{n \to \infty} \left(\frac{(n+1)(n+2) \cdots (2n)}{n^n}\right)^{1/n} \\[10pt]L = \lim_{n \to \infty} \left(\frac{n+1}{n} \cdot \frac{n+2}{n} \cdots \frac{n+n}{n}\right)^{1/n} \\[10pt]L = \lim_{n \to \infty} \left(\prod_{k=1}^n \left(1 + \frac{k}{n}\right)\right)^{1/n} \\[10pt]\text{Take the natural logarithm of both sides:} \\[10pt]\ln L = \lim_{n \to \infty} \frac{1}{n} \ln \left(\prod_{k=1}^n \left(1 + \frac{k}{n}\right)\right) \\[10pt]\ln L = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln \left(1 + \frac{k}{n}\right) \\[10pt]\text{This is a Riemann sum, which can be expressed as an integral:} \\[10pt]\ln L = \int_0^1 \ln(1+x) \, dx \\[10pt]\text{Now, we can integrate by parts:} \\[10pt]\text{Let } u = \ln(1+x) \text{ and } dv = dx. \\[10pt]\text{Then } du = \frac{1}{1+x} dx \text{ and } v = x. \\[10pt]\int_0^1 \ln(1+x) \, dx = \left[x \ln(1+x)\right]_0^1 - \int_0^1 \frac{x}{1+x} \, dx \\[10pt]= 1 \cdot \ln(2) - 0 - \int_0^1 \frac{x+1-1}{1+x} \, dx \\[10pt]= \ln 2 - \int_0^1 \left(1 - \frac{1}{1+x}\right) \, dx \\[10pt]= \ln 2 - \left[x - \ln(1+x)\right]_0^1 \\[10pt]= \ln 2 - \left[(1 - \ln 2) - (0 - \ln 1)\right] \\[10pt]= \ln 2 - (1 - \ln 2) \\[10pt]= 2 \ln 2 - 1 \\[10pt]= \ln 4 - 1 \\[10pt]= \ln 4 - \ln e \\[10pt]= \ln \left(\frac{4}{e}\right) \\[10pt]\text{Since } \ln L = \ln \left(\frac{4}{e}\right), \implies L = \frac{4}{e}$$​​