How do you find the limit of sqrt(x^2-9)/(2x-6) as x approaches -oo?

May 23, 2016

Do a little factoring to get ${\lim}_{x \to - \infty} = - \frac{1}{2}$.

Explanation:

When we deal with limits at infinity, it's always helpful to factor out an $x$, or an ${x}^{2}$, or whatever power of $x$ simplifies the problem. For this one, let's factor out an ${x}^{2}$ from the numerator and an $x$ from the denominator:
${\lim}_{x \to - \infty} \frac{\sqrt{{x}^{2} - 9}}{2 x - 6} = \frac{\sqrt{\left({x}^{2}\right) \left(1 - \frac{9}{{x}^{2}}\right)}}{x \left(2 - \frac{6}{x}\right)}$
$= \frac{\sqrt{{x}^{2}} \sqrt{1 - \frac{9}{{x}^{2}}}}{x \left(2 - \frac{6}{x}\right)}$

Here's where it begins to get interesting. For $x > 0$, $\sqrt{{x}^{2}}$ is positive; however, for $x < 0$, $\sqrt{{x}^{2}}$ is negative. In mathematical terms:
$\sqrt{{x}^{2}} = \left\mid x \right\mid$ for $x > 0$
$\sqrt{{x}^{2}} = - x$ for $x < 0$

Since we're dealing with a limit at negative infinity, $\sqrt{{x}^{2}}$ becomes $- x$:
$= \frac{- x \sqrt{1 - \frac{9}{{x}^{2}}}}{x \left(2 - \frac{6}{x}\right)}$
$= \frac{- \sqrt{1 - \frac{9}{{x}^{2}}}}{2 - \frac{6}{x}}$

Now we can see the beauty of this method: we have a $\frac{9}{x} ^ 2$ and $\frac{6}{x}$, both of which will go to $0$ as $x$ goes to negative infinity:
${\lim}_{x \to - \infty} = \frac{- \sqrt{1 - 0}}{2 - 0}$
${\lim}_{x \to - \infty} = - \frac{1}{2}$