<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] The key is to multiply by the conjugates of both radicals, in order to eliminate what makes the problem annoying: the two radicals whose limits both go to zero. And then you see what cancels out. 
$$\begin{align*}
\lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1}
\;&=\;
\lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl((3-x) - 1\bigr)}
&\qquad\qquad\tag{multiply by conjugates}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\bigl(2-x\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl(2 - x\bigr)}
&\qquad\qquad\tag{simplify}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\sqrt{3-x} + 1}{\sqrt{6-x} + 2}
&\qquad\qquad\tag{simplify more}
\\[2ex]&=\;
\frac{\lim_{x \to 2} \bigl(\sqrt{3-x} + 1\bigr)}{\lim_{x \to 2} \bigl(\sqrt{6-x} + 2\bigr)}
\tag{as both limits exist $\ldots$}
\\[2ex]&=\;
\frac{1 + 1}{2 + 2} \;=\; \frac{1}{2}.
\tag{$\ldots$ by substitution}
\end{align*}$$
</s>