Verification for a series of limits I ask kindly if my steps for these limits are right.
I start, always immediatly, checking the domain and this is an advice for my students of an high school.
I have, for example this function where the domain of
$$\operatorname{dom}\left(\frac{(1-x)^{\sqrt x}}{x-2}\right)=\{x\in \Bbb R \colon 0\leq x \leq 1\}$$
and I think that it is not possible to write that $x\to +\infty$
If I have the function $g(x)=\left(\frac{x-1}{x-2}\right)^{\sqrt x}$ where it is possibile to have $x\to +\infty$, my steps are:
$$\lim _{x\to +\infty }\left(\frac{x-1}{x-2}\right)^{\sqrt x}=\lim_{x\to +\infty}\left(1+\frac{1}{x-2}\right)^{\sqrt x}= \lim_{x\to +\infty}\left(1+\frac{1}{x-2}\right)^{\frac{(x-2)\sqrt x}{x-2}} $$
$$=\lim_{x\to +\infty}\left[\left(1+\frac{1}{x-2}\right)^{x-2}\right]^{\sqrt{\frac{x}{(x-2)^2}}}=e^0=1  $$
If the function is:
$$\psi(x)=\frac{(x-1)^{\sqrt x}}{x-2}$$
it is possible from the simple domain that  $x\to +\infty$,
$$\lim _{x\to +\infty }\frac{(x-1)^{\sqrt x}}{x-2}=\lim _{x\to +\infty }\frac{e^{\sqrt{x}\ln \left(x-1\right)}}{x-2}=\frac{e^{\lim _{x\to +\infty}x\sqrt{x}\,\frac{\ln (x-1)}{x}}}{x-2} \tag{A}$$
Being $$\lim _{x\to +\infty \:}\frac{\ln \left(x-1\right)}{x}=0 \tag{B}$$
I will have the exponent $x\sqrt{x}\,\frac{\ln (x-1)}{x}$ (in (A) last step) an indeterminate form $(\infty\cdot0)$. Is there a strategy without the use of the L'Hopital rule for the (A)?

Let's not use L'Hopital's rule in $(A)$.
$$\bbox[yellow,5px,border:2px solid red]{\frac{(x-1)^{\sqrt x}}{x-2}=\frac{x^{(\sqrt x-1)}(1-x^{-1})^{x(\sqrt x)^{-1}}}{1-2x^{-1}}=P(x)\cdot Q(x)}\tag{1}$$
where, $$P(x)=x^{(\sqrt x-1)}, \quad \text{and} \quad  Q(x)=\frac{(1-x^{-1})^{x(\sqrt x)^{-1}}}{1-2x^{-1}}$$
Note that $\lim_{x\to \infty} Q(x)=1$  ($\,\text{Numerator}$ of $Q(x)$ tends to $e^0=1$).
Assume on the contrary that limit in $(1)$ exists(finitely) as $x\to \infty$ and is equal to $L\in \mathbb R$.
Hence by limit rules, $P(x)=\dfrac{P(x)Q(x)}{Q(x)}\implies \lim_{x\to \infty} P(x)=L$, which is a contradiction (do you see why?).
Hence, the limit doesn't exist finitely.
