how do i find the value of lim of (2x-6) over (square of x - square root of 3) as x approaches 3 if limit exist.

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- Sep 6th 2008, 02:03 PMchicanala2290Limits
how do i find the value of lim of (2x-6) over (square of x - square root of 3) as x approaches 3 if limit exist.

- Sep 6th 2008, 02:10 PMKrizalid
$\displaystyle \frac{2x-6}{\sqrt{x}-\sqrt{3}}=\frac{2\left( \sqrt{x}+\sqrt{3} \right)\left( \sqrt{x}-\sqrt{3} \right)}{\sqrt{x}-\sqrt{3}}=2\left( \sqrt{x}+\sqrt{3} \right)\to 4\sqrt{3}$ as $\displaystyle x\to3.$

- Sep 6th 2008, 02:13 PMgalactus
This this what you mean?:

$\displaystyle \lim_{x\to 3}\frac{2x-6}{(x-\sqrt{3})^{2}}$

or is it:

$\displaystyle \lim_{x\to 3}\frac{2x-6}{x^{2}-\sqrt{3}}$

or none of the above?. - Sep 6th 2008, 02:57 PMchicanala2290
- Sep 6th 2008, 03:12 PMgalactus
You had 'square of x', not square root of x. Anyway, Kriz nailed it.