# help finding the limit..

• October 2nd 2008, 05:59 AM
Manizzle
help finding the limit..
i need to find the limit as x -> negative infinity for the function

(sqrt(x^2-2x))/(2x+7)
• October 2nd 2008, 06:27 AM
Chop Suey
Multiply by $\frac{\tfrac{1}{x}}{\tfrac{1}{x}}$ to get

$\frac{\sqrt{x^2-2x}}{2x+7} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}$

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

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