1. **Problem statement:** Calculate the limit $$\lim_{n \to \infty} x_n$$ where $$x_n = \left(\frac{(n!)^3}{n^{3n} e^{-n}}\right)^{\frac{1}{n}}.$$ We are asked to solve part (e) using the sandwich (squeeze) method. 2. **Rewrite the expression:** $$x_n = \left(\frac{(n!)^3}{n^{3n} e^{-n}}\right)^{\frac{1}{n}} = \frac{(n!)^{3/n}}{n^{3n/n} e^{-n/n}} = \frac{(n!)^{3/n}}{n^3 e^{-1}} = e \cdot \frac{(n!)^{3/n}}{n^3}.$$ 3. **Use Stirling's approximation:** For large $$n$$, $$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.$$ Taking the natural logarithm, $$\ln(n!) \approx n \ln n - n + \frac{1}{2} \ln(2 \pi n).$$ 4. **Calculate $$\ln x_n$$:** $$\ln x_n = 1 + \frac{3}{n} \ln(n!) - 3 \ln n.$$ Substitute the approximation: $$\ln x_n \approx 1 + \frac{3}{n} \left(n \ln n - n + \frac{1}{2} \ln(2 \pi n)\right) - 3 \ln n = 1 + 3 \ln n - 3 + \frac{3}{2n} \ln(2 \pi n) - 3 \ln n = 1 - 3 + \frac{3}{2n} \ln(2 \pi n) = -2 + \frac{3}{2n} \ln(2 \pi n).$$ 5. **Evaluate the limit:** As $$n \to \infty$$, $$\frac{3}{2n} \ln(2 \pi n) \to 0$$ because $$\ln n$$ grows slower than $$n$$. Therefore, $$\lim_{n \to \infty} \ln x_n = -2,$$ which implies $$\lim_{n \to \infty} x_n = e^{-2}.$$ 6. **Conclusion:** The limit is $$\boxed{e^{-2}}.$$