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.
Follow Math Help Forum on Facebook and Google+
$\displaystyle \frac{2x-6}{\sqrt{x}-\sqrt{3}}=\frac{2\left( \sqrt{x}+\sqrt{3} \right)\left( \sqrt{x}-\sqrt{3} \right)}{\sqrt{x}-\sqrt{3}}=2\left( \sqrt{x}+\sqrt{3} \right)\to 4\sqrt{3}$ as $\displaystyle x\to3.$
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?.
Originally Posted by galactus 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?. like the bottom one except that its not x^2 its square root of x
You had 'square of x', not square root of x. Anyway, Kriz nailed it.
View Tag Cloud