1. ## Compute the limit

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

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

What's the trick with this one?

2. $\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} = \frac{{(2 - x)\left( {\sqrt {3 - x} + 1} \right)}}{{(2 - x)\left( {\sqrt {6 - x} + 2} \right)}}$

3. Originally Posted by Plato
$\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} = \frac{{(2 - x)\left( {\sqrt {3 - x} + 1} \right)}}{{(2 - x)\left( {\sqrt {6 - x} + 2} \right)}}$
Thanks. Could you perhaps show me the steps you took to achieve that factorisation? Either numerator or denominator will do, I think

4. $2-x=\left( \sqrt{6-x}+2 \right)\left( \sqrt{6-x}-2 \right)=\left( \sqrt{3-x}+1 \right)\left( \sqrt{3-x}-1 \right).$

5. How are you guys getting all these fancy symbols? Do I not have enough posts yet or something?

6. It's using an inbuilt LaTeX compiler on this site.

For tutorials, see the LaTeX subforum.