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
 A: 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. 
$$\begin{align*}
\lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1}
\;&=\;
\lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl((3-x) - 1\bigr)}
&\qquad\qquad\tag{multiply by conjugates}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\bigl(2-x\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl(2 - x\bigr)}
&\qquad\qquad\tag{simplify}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\sqrt{3-x} + 1}{\sqrt{6-x} + 2}
&\qquad\qquad\tag{simplify more}
\\[2ex]&=\;
\frac{\lim_{x \to 2} \bigl(\sqrt{3-x} + 1\bigr)}{\lim_{x \to 2} \bigl(\sqrt{6-x} + 2\bigr)}
\tag{as both limits exist $\ldots$}
\\[2ex]&=\;
\frac{1 + 1}{2 + 2} \;=\; \frac{1}{2}.
\tag{$\ldots$ by substitution}
\end{align*}$$
A: Multiplying by conjugates works.  I suggest not getting discouraged by the complexity, because it works out in the end.
An alternative is to rewrite it as
$$
\left(\frac{\sqrt{6-x}-2}{x-2}\right) \cdot \frac{1}{\left(\dfrac{\sqrt{3-x}-1}{x-2}\right)}$$
and notice that each parenthesized expression has the form $\dfrac{f(x)-f(2)}{x-2}$ (for two different $f$s).  Then you can find the limit by evaluating the derivatives, i.e., limits of difference quotients, and using the properties of limits of products and reciprocals.
A: 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 $(1)$:
$$
\begin{align}
\lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&=\lim_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\
&=\frac12\tag{2}
\end{align}
$$
A: You are right to try conjugates:  $\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}=\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}{{\sqrt{3-x}+1}\over {\sqrt{3-x}+1}}=\lim_{x\to2}\frac{\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2}{2-x}$ which is still $\frac 00$, but we can define $u=2-x$ and try a Taylor series
$\lim_{x\to2}\frac{\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2}{2-x}=\lim_{u \to 0} \frac{\sqrt{(4+u)(1+u)}+\sqrt{4+u}-2\sqrt{1+u}-2}{u}=\lim_{u \to 0}\sqrt{1+\frac 5u+4}+\sqrt{\frac4{u^2}+\frac 1u}-\sqrt{\frac 4{u^2}+\frac 4u}-\frac 2u$
Now pull out the terms in $\frac 1u$, which should cancel and you will be left with something finite.
