# Thread: Limit as X approaches negative infinity

1. ## Limit as X approaches negative infinity

i need to find the limit as x -> negative infinity for the function

(sqrt(x^2-2x))/(2x+7)

2. The limit is -1/2. $\displaystyle \sqrt{x^2 - 2x} : \sqrt{x^2 - 2x + 1} = \sqrt{(x-1)^2} = |x-1|$, and $\displaystyle \frac{|x-1|}{2x + 7} : \frac{|x|}{2x}$. Here I am using ":" to mean "approximately equal."

http://www.mathhelpforum.com/math-he...ing-limit.html

4. yes but i needed the first answer and didn't know how to ask for it differently

5. He wasn't wrong, Both methods are correct.

6. Actually, I did make a mistake and I'm sorry about that. Regardless, you could've still pointed it out in the original thread.

The problem is when I multiplied $\displaystyle \sqrt{x^2 -2x}$ by $\displaystyle \frac{1}{x}$. I'll leave it to you to see what went wrong.

An alternative way to do this is to take x^2 and x common factor:
$\displaystyle \frac{\sqrt{x^2-2x}}{2x+7}$

$\displaystyle = \frac{\sqrt{x^2(1-\frac{2}{x})}}{x(2+\frac{7}{x})}$

$\displaystyle = \frac{|x|\sqrt{1-\frac{2}{x}}}{x(2+\frac{7}{x})}$

Since limit goes to negative infinity, absolute value of x is turned into -x. And I'm sure you can follow from there.