i need to find the limit as x -> negative infinity for the function
(sqrt(x^2-2x))/(2x+7)
You have already asked this before here:
http://www.mathhelpforum.com/math-he...ing-limit.html
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.