Compute the limit

**rowe** · January 8th 2010, 07:44 AM

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?

**Plato** · January 8th 2010, 08:05 AM

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

**rowe** · January 8th 2010, 08:21 AM

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

**Krizalid** · January 8th 2010, 08:25 AM

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

**BugzLooney** · January 8th 2010, 08:25 PM

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

**Prove It** · January 8th 2010, 09:18 PM

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

For tutorials, see the LaTeX subforum.

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