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?

Just another way:
Let $\sqrt{6-x}-2=h,\sqrt{3-x}-1=k\implies h\to0,k\to0$
and $$6-4-4h-h^2=x=3-1-2k-k^2\implies-h(4+h)=-k(k+2)\implies\dfrac hk=\dfrac{k+2}{h+4}$$
$$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{h\to0,k\to0}\dfrac hk=\lim_{h\to0,k\to0}\dfrac{k+2}{k+4}=?$$
