Question about limits at infinity with radicals.

Kstanley

1. The problem statement, all variables and given/known data
Find the following limit:
[tex]
\lim_{x \to \infty} \frac{2+\sqrt{(6x)}}{-2+\sqrt{(3x)}}
[/tex]

2. Relevant equations
n/a


3. 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!

Mark44

I would factor [itex]\sqrt{x}[/itex] out of both terms in the numerator and both in the denominator.

That would give you
[tex]\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x}}\cdot \frac{2/\sqrt{x}+\sqrt{6}}{-2/\sqrt{x}+\sqrt{3}}[/tex]

ystael

No, you can't do it this way. You can say the square roots "go to zero" and can be cancelled out, but you are right to observe that by the same argument, the [tex]2/x[/tex] terms should "go to zero" as well.

Try multiplying the numerator and denominator by [tex]x^{-1/2}[/tex] rather than [tex]x^{-1}[/tex].

Kstanley

Ok I understand multiplying the numerator and denominator by 1/sqrt(X) to obtain:

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

From this point do I multiply by the conjugate of the numerator?

[tex]
\lim_{x \to \infty} \frac{(\frac{2}{\sqrt{x}})^2-{6}}{-(\frac{2}{\sqrt{x}})^2+\frac{2\sqrt{3}}{\sqrt{x}}+\frac{2\sqrt{6}}{\sqr t{x}}-\sqrt{18}}
[/tex]

This seems like quite the jumbled mess so maybe I'm wrong..

Mark44

No, just take the limit. The first terms in the top and bottom go to zero.

Kstanley

Ah duh I got so into thinking it had to be complicated I didn't even think to just find the limit right there. Thank you so much that was so much easier than I imagined!