- #1

Kstanley

- 7

- 0

## Homework Statement

Find the following limit:

[tex]

\lim_{x \to \infty} \frac{2+\sqrt{(6x)}}{-2+\sqrt{(3x)}}

[/tex]

## Homework Equations

n/a

## The Attempt at a Solution

I know this shouldn't be that hard, but somehow I keep getting stuck on simplifying the equation. I think the first step is to multiply both the denominator and numerator by 1/x (which equals 1/sqrt(x^2). This would give me

[tex] \lim_{x \to \infty} \frac{\frac{2}{x}+\sqrt{\frac{6x}{x^2}}}{\frac{-2}{x}+\sqrt{\frac{3x}{x^2}}} [/tex]

And if I'm correct the squareroots should go to zero which leaves:

[tex] \lim_{x \to \infty} \frac{\frac{2}{x}}{\frac{-2}{x}} [/tex]

But wouldn't those on top of eachother equal 0/0 too unless I can multiply by x for both so it equals 2/-2?

I am really just not sure of the correct way to go about simplifying this. Any help would be appreciated!