Answer
$$
\lim _{x \rightarrow 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{1}{2}
$$
Work Step by Step
$$
\begin{aligned}
\lim _{x \rightarrow 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\lim _{x \rightarrow 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \frac{\sqrt{6-x}+2}{\sqrt{6-x}+2} \cdot \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\right)\\
&=\lim _{x \rightarrow 2}\left[\frac{(\sqrt{6-x})^{2}-2^{2}}{(\sqrt{3-x})^{2}-1^{2}} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right]\\
&=\lim _{x \rightarrow 2}\left(\frac{6-x-4}{3-x-1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)\\
&=\lim _{x \rightarrow 2} \frac{(2-x)(\sqrt{3-x}+1)}{(2-x)(\sqrt{6-x}+2)}\\
&=\lim _{x \rightarrow 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\
&=\frac{1}{2}
\end{aligned}
$$
