91Ó°ÊÓ

Take the natural logarithm of the sequence: \begin{align*} \ln\left(\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}\right) &= \frac{1}{n}\ln\left(\frac{n!}{(mn)^n}\right). \end{align*}

Stirling's approximation states that \(n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\) as n approaches infinity. Replace \(n!\) with the approximation: $$\frac{1}{n}\ln\left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{(mn)^n}\right).$$

Simplify the expression inside the logarithm and then split the logarithm into the sum of logarithms. The simplified expression will be: \begin{align*} \frac{1}{n}\left[\ln\left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{(mn)^n}\right)\right] &= \frac{1}{n} \left[\frac{1}{2}\ln(2\pi n) + n\ln\left(\frac{n}{e}\right) - n\ln(mn)\right]. \end{align*}

Distribute \(\frac{1}{n}\) inside the brackets and then find the limit as n approaches infinity. \begin{align*} \lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n) + \ln\left(\frac{n}{e}\right) - \ln(mn)\right] &= \lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n)\right] + \lim_{n\to\infty} \left[\ln\left(\frac{n}{e}\right)\right] - \lim_{n\to\infty} \left[\ln(mn)\right]. \end{align*} Now, notice that \(\lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n)\right] = 0\) since the logarithm grows slower than the linear term in the denominator. And, since m is a constant, we have \(\lim_{n\to\infty} \left[\ln(mn)\right] = \ln(m)\). So, the limit becomes: $$\lim_{n\to\infty} \left[\ln\left(\frac{n}{e}\right) - \ln(m)\right].$$

Combine the logarithms: $$\lim_{n\to\infty} \left[\ln\left(\frac{n}{me}\right)\right].$$

Take the exponential of both sides to reverse the process of taking the logarithm: $$\lim_{n\to\infty}\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} e^{\left[\ln\left(\frac{n}{me}\right)\right]} = \frac{1}{m e}.$$ Hence, the correct answer is (A) \(\frac{1}{em}\).