Compute $\displaystyle \lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$.

I approached this by multiplying the whole expression by the conjugate of the numerator, $\displaystyle \sqrt{6-x}+2$, which gave me a non-zero denominator, but evaluated the limit as 0. The actual limit is $\displaystyle \tfrac{1}{2}$.

What's the trick with this one?