# Math Help - Negative limits

1. ## Negative limits

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.

2. Originally Posted by unluckykc
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 $x \to - \infty$, the 1 in the top and the $x$ in the bottom do not really make a difference, so we can get rid of them.

thus, $\lim_{x \to - \infty} \frac {-3x + 1}{\sqrt{x^2 + x}} = \lim_{x \to - \infty} \frac {-3x}{\sqrt{x^2}} = \lim_{x \to - \infty} \frac {-3x}{|x|} = \left \{ \begin{array}{lr} -3 & \mbox{ as } x \to \infty \\ 3 & \mbox{ as } x \to -\infty \end{array}\right.$

3. Since $\sqrt{x^{2}}=|x|$ we can divide by -x and $\sqrt{x^{2}}$ where we need to.

$\lim_{x\rightarrow{-\infty}}\frac{\frac{-3x}{-x}+\frac{1}{-x}}{\sqrt{1+\frac{1}{x}}}$

Now, see the limit?.

4. and again, we could split the original into two pieces.

$\frac {-3x + 1}{\sqrt{x^2 + x}} = \frac {-3x }{\sqrt{x^2 + x}} + \frac 1{\sqrt{x^2 + x}}$

now the second fraction goes to zero as $x \to - \infty$ so we can forget about it. the first fraction we can simplify into $\frac {-3x}{|x|}$ as $x \to - \infty$ and the conclusion follows