<s>[INST] Limit of quotients with square roots: $\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ You can't use L'Hospital's rule :S
$$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
 [/INST] This is not any different than some of the other solutions, just hopefully a bit easier to follow:
$$
\begin{align}
\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&=\frac{\color{#C00000}{\sqrt{6-x}-2}}{\color{#00A000}{\sqrt{3-x}-1}}
\frac{\color{#C00000}{\sqrt{6-x}+2}}{\color{#C00000}{\sqrt{6-x}+2}}
\frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#00A000}{\sqrt{3-x}+1}}\\
&=\color{#C00000}{\frac{(6-x)-4}{\sqrt{6-x}+2}}
\color{#00A000}{\frac{\sqrt{3-x}+1}{(3-x)-1}}\\
&=\frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#C00000}{\sqrt{6-x}+2}}
\frac{\color{#C00000}{2-x}}{\color{#00A000}{2-x}}\\
&=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\tag{1}
\end{align}
$$
Take $\lim\limits_{x\to2}$ of $(1)$:
$$
\begin{align}
\lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&=\lim_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\
&=\frac12\tag{2}
\end{align}
$$
</s>