Lim (x>-infinity) (-3x + 1)/(x^2 + x)^(1/2).
I think it should be 3, but I don't know how to show it.
as $\displaystyle x \to - \infty$, the 1 in the top and the $\displaystyle x$ in the bottom do not really make a difference, so we can get rid of them.
thus, $\displaystyle \lim_{x \to - \infty} \frac {-3x + 1}{\sqrt{x^2 + x}} = \lim_{x \to - \infty} \frac {-3x}{\sqrt{x^2}} = \lim_{x \to - \infty} \frac {-3x}{|x|} = \left \{ \begin{array}{lr} -3 & \mbox{ as } x \to \infty \\ 3 & \mbox{ as } x \to -\infty \end{array}\right.$
and again, we could split the original into two pieces.
$\displaystyle \frac {-3x + 1}{\sqrt{x^2 + x}} = \frac {-3x }{\sqrt{x^2 + x}} + \frac 1{\sqrt{x^2 + x}}$
now the second fraction goes to zero as $\displaystyle x \to - \infty$ so we can forget about it. the first fraction we can simplify into $\displaystyle \frac {-3x}{|x|}$ as $\displaystyle x \to - \infty$ and the conclusion follows