# Math Help - Limit as X approaches infinity

1. ## Limit as X approaches infinity

Hi,
I was given some review questions for a test and one of them is
Lim as(x->infinity) of (2x-(4x^2+2x)^1/2). I've tried multiplying by the reciprocal but it didn't seem to get me anywhere; I'm not really sure how to approach the problem in any other way. Is there something I'm missing?
Thanks!

2. Originally Posted by geliot
Hi,
I was given some review questions for a test and one of them is
Lim as(x->infinity) of (2x-(4x^2+2x)^1/2). I've tried multiplying by the reciprocal
Did you get $\dfrac{-2x}{2x+\sqrt{4x^2+2x}}?$

You should have. Then divide through by $2x$.

3. Originally Posted by Plato
Did you get $\dfrac{-2x}{2x+\sqrt{4x^2+2x}}?$

You should have. Then divide through by $2x$.
oh wow I'm feeling pretty dumb right now. I forgot to square the 2x >.>
Thanks so much!

4. but sorry actually, now that I'm looking at it, how do I pull the 2x out of the radical? Looking at the answer, the limit approaches -.5, but I don't see how one would end up with a 2 on the bottom of the fraction...sorry, I'm sure i'm just missing something obvious here

5. Originally Posted by geliot
but sorry actually, now that I'm looking at it, how do I pull the 2x out of the radical? Looking at the answer, the limit approaches -.5, but I don't see how one would end up with a 2 on the bottom of the fraction...sorry, I'm sure i'm just missing something obvious here
$\displaystyle \frac{-2x}{2x+\sqrt{4x^2+2x}} = \frac{-2}{2 + \frac{\sqrt{4x^2+2x}}{x}} = \frac{-2}{2 + \sqrt{\frac{4x^2+2x}{x^2}}} = ....$