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.


Option:
$z:=\sqrt{3-x}$; and consider $\lim z \rightarrow 1;$
$\dfrac{\sqrt{3+z^2}-2}{z-1}=$
$\dfrac{z^2-1}{(z-1)(\sqrt{3+z^2}+2)}=$
$\dfrac{z+1}{\sqrt{3+z^2}+2}.$
Take the limit.
