Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1}   \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$

Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.


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.
