i need to find the limit as x -> negative infinity for the function
(sqrt(x^2-2x))/(2x+7)
Multiply by $\displaystyle \frac{\tfrac{1}{x}}{\tfrac{1}{x}}$ to get
$\displaystyle \frac{\sqrt{x^2-2x}}{2x+7} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}}$
$\displaystyle = \frac{\sqrt{\frac{x^2-2x}{x^2}}}{2 + \frac{7}{x}} $
$\displaystyle = \frac{\sqrt{1-\frac{2}{x}}}{2 + \frac{7}{x}}$