<s>[INST] How can I find this limit that is supposed to converge to e? I made a formula to get an approximation $x-1$ based on $x$; $f(x) = \frac{(-2 + x)^{3/2 - x} x^{-1/2 + x}}{e^2}$. I got this formula by looking at $\frac{n!}{n(n-2)!}$ which equals $n-1$, then substituting the factorials with Stirling approximants. After simplifying through Wolfram Alpha, I got my function. Then, I realized I could solve for $e$, and I got $$\lim_{x \rightarrow \infty} \frac{(x - 2)^{3/4} \sqrt{x^x}}{\sqrt{{(x - 2)^x}{(x - 1)}{\sqrt{x}}}} = e$$. The problem is, I cannot figure out how to solve this limit. It should converge to e (albeit very slowly) but I do not know how to make progress or even simplify this limit significantly. I want to know, if I were presented this limit without knowing how it was obtained, how I could solve it.
 [/INST] $$ f(x)=\frac{(x-2)^{\frac{3}{2}-x} x^{x-\frac{1}{2}}}{e^2}\implies \log[f(x)]=\left(\frac{3}{2}-x\right)\log(x-2)-\left(x-\frac{1}{2}\right)\log(x)-2$$ Expanding as series for large values of $x$
$$\log[f(x)]=\log (x)-\frac{1}{x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^4}\right)$$
$$f(x)=e^{\log[f(x)]}=x-1+\frac{1}{6 x}+\frac{1}{6 x^2}+O\left(\frac{1}{x^3}\right)$$
Edit
For the second one
$$g(x)= \frac{(x - 2)^{3/4} \sqrt{x^x}}{\sqrt{{(x - 2)^x}{(x - 1)}{\sqrt{x}}}} \implies g^2(x)=\frac{(x-2)^{\frac{3}{2}-x} x^{x-\frac{1}{2}}}{x-1}$$
$$\log[g^2(x)]=\left(\frac{3}{2}-x\right) \log (x-2)+\left(x-\frac{1}{2}\right)\log(x)-\log(x-1)$$
$$\log[g^2(x)]=2+\frac{1}{6 x^2}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$\log[g(x)]=1+\frac{1}{12 x^2}+\frac{1}{6 x^3}+O\left(\frac{1}{x^4}\right)$$
$$g(x)=e^{\log[g(x)]}=e \Big[1+\frac{1}{12 x^2}+\frac{1}{6 x^3} \Big]+O\left(\frac{1}{x^4}\right)$$
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