Calculus: Early Transcendentals, Chapter 2, Review Exercises, Section Review Exercises, Problem 14

Determine the limit $\displaystyle \lim_{x \to -\infty} \frac{\sqrt{x^2 -9 }}{2x - 6}$

$\displaystyle \lim_{x \to -\infty} = \frac{\sqrt{x^2 - 9}}{2x - 6} = \frac{\sqrt{(x + 3)(x - 3)}}{2(x - 3)}$

By dividing both numerator and denominator by $x$ and by using the property of limits, we have

$
\begin{equation}
\begin{aligned}
\lim_{x \to -\infty} \frac{\sqrt{x^2 - 9}}{2x - 6} &= \lim_{x \to -\infty} \frac{\sqrt{1 - \frac{9}{x^2}}}{2 - \frac{6}{x}}
&& \text{Since $\sqrt{x^2} = x$ for } x > 0\\
\\
&= \frac{\displaystyle \lim_{x \to -\infty}\sqrt{1 - \frac{9}{x^2}}}{\displaystyle \lim_{x \to -\infty} \left( 2 - \frac{6}{x} \right)}
= \frac{\sqrt{\displaystyle \lim_{x \to -\infty} 1 - 9 \lim_{x \to -\infty} \frac{1}{x^2} }}{\displaystyle \lim_{x \to -\infty} 2 - 6 \lim_{x \to -\infty} \frac{1}{x}}\\
\\
&= \frac{\sqrt{1 - 9 \cdot 0}}{2 - 6 \cdot 0} = \frac{\sqrt{1}}{2} = \frac{1}{2}
\end{aligned}
\end{equation}
$


In computing the limit as $x \to - \infty$, we must remember that for $x < 0$, we have $\sqrt{x^2} = |x| = - x$.
So when we divide the numerator by $x$, for $x < 0$ we get
$\displaystyle \frac{1}{x} \sqrt{x^2 - 9} = -\frac{1}{\sqrt{x^2}} \sqrt{x^2 - 9} = -\sqrt{1 - \frac{9}{x^2}}$

Therefore,

$
\begin{equation}
\begin{aligned}
\lim_{x \to -\infty} \frac{\sqrt{x^2 - 9}}{2x - 6} &= \lim_{x \to -\infty} \frac{-\sqrt{1 - \frac{9}{x^2}}}{2 - \frac{6}{x}}\\
\\
&= \frac{-\sqrt{\displaystyle 1 - 9 \lim_{x \to -\infty} \frac{1}{x^2} }}{\displaystyle 2 - 6 \lim_{x \to -\infty} \frac{1}{x}}\\
\\
&= \frac{-\sqrt{1 - 9 \cdot 0}}{2 - 6 \cdot 0}\\
\\
&= \frac{-\sqrt{1}}{2} = \frac{-1}{2}
\end{aligned}
\end{equation}
$

Thus,
$\displaystyle \lim_{x \to -\infty} \frac{\sqrt{x^2 - 9 }}{2x -6} = -\frac{1}{2}$