1. **State the problem:** Find the limit as $x \to \infty$ of $$8e^{\sqrt{\frac{x}{x+1}}} \left(\frac{x-1}{x}\right)! - \left(8x^2 - 4x\ln x - \ln 2x - \frac{4x + 2\ln x}{\ln 2\pi}\right).$$ 2. **Analyze the factorial term:** The term $\left(\frac{x-1}{x}\right)!$ is a factorial of a number approaching 1 from below as $x \to \infty$ because $$\frac{x-1}{x} = 1 - \frac{1}{x} \to 1.$$ Factorials for non-integers are defined by the Gamma function: $$n! = \Gamma(n+1).$$ So, $$\left(\frac{x-1}{x}\right)! = \Gamma\left(\frac{x-1}{x} + 1\right) = \Gamma\left(2 - \frac{1}{x}\right).$$ As $x \to \infty$, this approaches $\Gamma(2) = 1! = 1$. 3. **Evaluate the exponential term:** $$\sqrt{\frac{x}{x+1}} = \sqrt{1 - \frac{1}{x+1}} \to 1$$ as $x \to \infty$. Thus, $$e^{\sqrt{\frac{x}{x+1}}} \to e^1 = e.$$ 4. **Combine the first term:** $$8e^{\sqrt{\frac{x}{x+1}}} \left(\frac{x-1}{x}\right)! \to 8e \times 1 = 8e.$$ 5. **Analyze the second term:** $$8x^2 - 4x\ln x - \ln 2x - \frac{4x + 2\ln x}{\ln 2\pi}.$$ As $x \to \infty$, the dominant term is $8x^2$, which grows without bound. 6. **Conclusion:** The first term tends to a finite constant $8e$, but the second term grows like $8x^2$ which dominates and tends to infinity. Therefore, $$\lim_{x \to \infty} \left[8e^{\sqrt{\frac{x}{x+1}}} \left(\frac{x-1}{x}\right)! - \left(8x^2 - 4x\ln x - \ln 2x - \frac{4x + 2\ln x}{\ln 2\pi}\right)\right] = -\infty.$$