# Need help over a problem with horizontal assymptote

• Dec 3rd 2010, 10:37 PM
Kakariki
Need help over a problem with horizontal assymptote
Hey,
Basically I have attempted a solution to my problem, it did not get the correct answer and I cannot figure out how to get the correct answer. Looking for where I am making my mistake and how to correct it!

Question
Find the limit
$\lim_{x\to -\infty} (x + \sqrt{x^2 + 2x})$

Attempted solution:
*pretend that $\lim_{x\to -\infty}$ is in front, I am making a mess trying to coordinate this with that and it's just easier to show without..

$(x + \sqrt{x^2 + 2x})* \frac{(x - \sqrt{x^2 + 2x})}{(x - \sqrt{x^2 + 2x})}$ ***Multiplied by conjugate
$\frac{-2x}{x - \sqrt{x^2 + 2x}}$
The highest value of x is "x" so I divide that through the numerator and denominator and get:
$\frac{-2}{1 - \sqrt{1}}$
Which is giving me 0 on the bottom. I do not know how to fix this and was hoping someone could give me a hand here.

Thanks alot!
• Dec 3rd 2010, 10:46 PM
chisigma
Take into account that is...

$\displaystyle \lim_{x \rightarrow - \infty} x + \sqrt{x^{2} + 2 x} = \lim_{x \rightarrow + \infty} \sqrt{x^{2} - 2 x} - x$

Kind regards

$\chi$ $\sigma$