<s>[INST] 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.
 [/INST] 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*}
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