Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e.   $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}$$ and conversely $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
To my suspicions of which–either numerator or denominator–conjugate to multiply by I chose $$\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
This resulted in $$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{3-x-1}$$
Is it indeterminate? What is the reason for multiplying by a specific conjugate in a fraction (denominator or numerator) and the reason for the conjugate being either a) denominator b) numerator c) or both?
Am I simply practicing incorrect algebra by rationalizing the expression to:
$$\frac{(6-x)(3-x)-2\sqrt{3}+2x+2}{x-2}$$
Or am I failing to delve further and manipulate the expression out of the indeterminate form?

By your idea 
$$\lim_{x\rightarrow2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(6-x-4)}{(\sqrt{6-x}+2)(3-x-1)}$$
$$=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(2-x)}{(\sqrt{6-x}+2)(2-x)}=\lim_{x\rightarrow2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{2}{4}=\frac{1}{2}$$
