Calculate $\lim \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1}$ as $x \to 2$ I don't know the answer to the problem.  None of my "tricks" work: I see no way of factoring it, rationalization on the denominator doesn't seem useful (because I still end up with a zero).  The book has not gone over any fancier techniques such as L'Hospital's rule.
Reference.  This problem comes from section $2.3$ of Stewart's Calculus.  It's problem $60$, an even number, so I don't even know the final answer.
What have I learned?  Sometimes we need to rationalize both numerator and denominator.
 A: $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{(6-x)-2^2}{(3-x)-1^2}\cdot\frac{\sqrt{3-x}+2}{\sqrt{6-x}+2}=\frac{2-x}{2-x}\cdot\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$$
Could you complete now?
A: $$\lim_{x\to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1}=\lim_{x\to 2}\frac{(\sqrt{6 - x} - 2)(\sqrt{3 - x} + 1)}{(\sqrt{3 - x} - 1)(\sqrt{3 - x} + 1)}=\lim_{x\to 2}\frac{(\sqrt{6 - x} - 2)(\sqrt{3 - x} + 1)}{2-x}$$
Multiply for the conjugate of $\sqrt{6 - x} - 2$,
$$=\lim _{x\to \:2}\left(\frac{\frac{-x+2}{\sqrt{6-x}+2}\left(\sqrt{3-x}+1\right)}{-x+2}\right)=\lim _{x\to \:2}\left(\frac{\sqrt{-x+3}+1}{\sqrt{-x+6}+2}\right)=\frac{\sqrt{-2+3}+1}{\sqrt{-2+6}+2}=\frac 12$$
A: The limit has the indeterminate form $0/0$, which can be eliminated if you rationalize both the numerator and denominator.
\begin{align*}
\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} \cdot \frac{\sqrt{6 - x} + 2}{\sqrt{6 - x} + 2} \cdot \frac{\sqrt{3 - x} + 1}{\sqrt{3 - x} + 1}\\
& = \lim_{x \to 2} \frac{(2 - x)(\sqrt{3 - x} + 1)}{(2 - x)(\sqrt{6 - x} + 2)}\\
& = \lim_{x \to 2} \frac{\sqrt{3 - x} + 1}{\sqrt{6 - x} + 2}\\
& = \frac{1}{2}
\end{align*}
