Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

Rationalization is a standard way to manipulate such kind of limits when they lead to an indeterminate form.

The aim for this kind of manipulation is to eliminate the term which leads to the indetermination, indeed by $(A-B)(A+B)=A^2-B^2 \implies A-B= \frac{A^2-B^2}{A+B}$ we have that

$$\sqrt{6-x}-2=\frac{2-x}{\sqrt{6-x}+2}$$

$$\sqrt{3-x}-1=\frac{2-x}{\sqrt{3-x}+1}$$

and by the ratio the problematic $x-2$ term cancel out.

As an alternative we can also use binomial first order approximation (i.e. Taylor's series) at $x=2$ to obtain

$$\sqrt{6-x}=\sqrt{4-(x-2)}=2\sqrt{1-\frac{(x-2)}4}=2\left(1-\frac {x-2}{8}+o(x-2)\right)$$

$$\implies \sqrt{6-x}-2=-\frac {x-2}{4}+o(x-2)$$

$$\sqrt{3-x}=\sqrt{1-(x-2)}=1-\frac {x-2}{2}+o(x-2)$$

$$\implies \sqrt{3-x}-1=-\frac {x-2}{2}+o(x-2)$$

which gives evidence of the same problematic term and the final result.

As written, no simplification is apparent, and this is due to the presence of the radicals. Now considering the identity

$$a-b=\frac{a^2-b^2}{a+b}$$ there is a hope of getting rid of them. In the case of your numerator,

$$\sqrt{6-x}-2=\frac{(\sqrt{6-x})^2-2^2}{\sqrt{6-x}+2}=\frac{2-x}{\sqrt{6-x}+2}.$$

Now the radical is gone from the numerator and has moved to the denominator, but it is important to notice that it does not cancel because the minus has turned to a plus.

Repeating this trick with the denominator of the original ratio, you will see a simplification.

If you replace with $t:= 2-x$, the question becomes

$$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1}$$

From wolframalpha, reading the diagram for $\sqrt{t+1}-1$, at the neighbour of $0$ it's something like $t$, or $$\sqrt{t+1}-1=t+O(t^2)$$; similarly, $$\sqrt{t+4}-2=\frac12 t+O(t^2)$$ , here we are using the big O notation.

Then $$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1} = \lim _{t\to 0}\frac{\frac12 t+O(t^2)}{t+O(t^2)}=\frac12$$

Notice the last step is reduction of a fraction $t$.

In your first result,

$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indetermination $\frac{0}{0}$

It's actually

$$\lim _{t\to 0} \left(\dfrac{(\sqrt{t+4}-2)(\sqrt{t+1}+1)}{t}\right) $$

, as the reduction hasn't been done, it's still $\frac{0}{0}$.

So the trick is to derive something as $t$ and reduce.

Sometimes, the problem is a bit tricky that need to do more than one round.

ps. in analysis, after limit, derivative would be explained, then its Taylor's expansion, at that time $\sqrt{t+1}-1=\frac12 t+O(t^2)$ will be more obvious.