In the spirit of paw88789's answer, first show:
$\color{red}{ 1 = \lim_{x\to\infty} 1 = \lim_{x\to\infty} \dfrac{x}{\sqrt{x}\sqrt{x}} \overset{???}{=} \dfrac{\infty}{\infty\infty} \overset{???}{=} \dfrac{1}{\infty} = 0 }$.
And ask the students what is wrong with it, and attempt to argue with them that this reasoning is correct. By forcing them to pinpoint the problem on their own, you automatically make them much better at avoiding the same erroneous thought process themselves.
Almost surely students will be able to 'point out' that the "$\infty$" on top is bigger than each of the individual "$\infty$"s below. Ignoring that $\infty$ isn't well-defined, pursue the matter and ask them, why so, and why should it make any difference? They should say that it was "$x$" on top but "$\sqrt{x}$" below, and that if you don't distinguish the 'infinities' then you get the incorrect conclusion that the top is 'smaller' than the bottom.
Then it is easy to now tell them that using a single symbol "$\infty$" has resulted in loss of information of how big $x$ is getting, so we simply cannot use "$\infty$". This should be enough to convince students of the futility in using the first 'method' in your example.
As for the correct solution, I do not recommend the one you propose but rather:
$\dfrac{x+2}{x(x-1)} = \dfrac{x(1+\frac{2}{x})}{x^2(1-\frac{1}{x})} = \dfrac{1}{x} \dfrac{1+\frac{2}{x}}{1-\frac{1}{x}} \to 0$ as $x \to \infty$.
With the following thought process:
Identify the significant terms in each subexpression. On the top, for large $x$ the $2$ is comparatively insignificant to the $x$, and at the bottom likewise. We want to see how fast it grows, so we isolate that main growth rate. On top it is $x$, and at the bottom it is $x^2$. As we can see, at the bottom the leftover is $1$ plus/minus something small, so indeed $x^2$ has the same significant part (digits) as the original denominator. After simplification, we are left with $\frac{1}{x}$ multiplied by something close to $1$ then divided by something close to $1$. Intuitively it is clear that the overall growth is significantly the same as $\frac{1}{x}$.
Why is this better? Because students get a real feel for the actual growth rate rather than attempting to blindly churn out the answer. Not only that, it generalizes readily without difficulty to asymptotic expansions, which are widely used in engineering, physics and chemistry but that students are hardly taught to be able to manipulate!
After practicing with many examples, it would be a good idea to show students the following error:
$\displaystyle \color{red}{ 1 = \lim_{n\to\infty} 1 = \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n} \overset{???}{=} \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\infty} = \lim_{n\to\infty} \sum_{k=1}^n 0 = \lim_{n\to\infty} 0 = 0 }$.
Again as before, it is best to draw the students into a discussion and try to 'convince' them that it is correct. It should become clear that not only is $\infty$ dangerous to try using, it is still very dangerous even if only the limiting variable is substituted by $\infty$! This idea is a very common but fatal one, as the following examples will show any students that remain stubbornly unconvinced that $\infty$ is not to be trifled with:
$\displaystyle \color{red}{ \lim_{n\to\infty} \sum_{k=1}^n \frac{k}{n^2} \overset{???}{=} \sum_{k=1}^\infty \frac{k}{\infty^2} \overset{???}{=} {???} }$.
And if they argue that $k$ is on average $\frac{1}{2} \infty$ in the above summation then show them:
$\displaystyle \color{red}{ \lim_{n\to\infty} \sum_{k=1}^n \frac{k^2}{n^3} \overset{???}{=} \sum_{k=1}^\infty \frac{(\frac{1}{2}\infty)^2}{\infty^2} \overset{???}{=} \infty \times \frac{\infty^2}{4\infty^3} \overset{???}{=} \frac{1}{4} }$. [FALSE!!!]