Limit as X approaches negative infinity

**Manizzle** · October 2nd 2008, 10:24 AM

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

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

**icemanfan** · October 2nd 2008, 10:26 AM

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

**Chop Suey** · October 2nd 2008, 10:36 AM

You have already asked this before here:
http://www.mathhelpforum.com/math-he...ing-limit.html

**Manizzle** · October 2nd 2008, 10:37 AM

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

**kbartlett** · October 2nd 2008, 10:47 AM

He wasn't wrong, Both methods are correct.

**Chop Suey** · October 2nd 2008, 11:38 AM

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 $\sqrt{x^2 -2x}$ by $\frac{1}{x}$ . 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:
$\frac{\sqrt{x^2-2x}}{2x+7}$

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

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

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

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