# Math Help - help finding the limit..

1. ## help finding the limit..

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

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

2. Multiply by $\frac{\tfrac{1}{x}}{\tfrac{1}{x}}$ to get

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

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

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