Strange limit problem to be solved without Hospital's Rule...? Having trouble solving this limit problem without L'Hôpital's Rule...

$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

Tried multiplying the function by the conjugate/inverse-conjugate, of both the numerator and denominator... but no avail.... any ideas?

$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{3-x}+1)}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$$
$$=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}=\lim_{x \to 2} \frac{(\sqrt{3-x}+1)(2-x)}{(2-x)(\sqrt{6-x}+2)}=$$
$$\lim_{x \to 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{1}{2}$$
