Can someone please explain to me how to solve $\lim_{x \to -\infty} \frac {\sqrt{x^2 - 9}}{2x - 6}$ Can someone please explain to me how to solve $\lim_{x \to -\infty} \frac {\sqrt{x^2 - 9}}{2x - 6}$
In the following solution the only thing that I don't understand is the appearance of that minus sign before the limit after we include the divisor under root sign.
solution on wolframalpha
 A: When you push a positive number inside a square root, it squares itself:
$$3\sqrt{x+2} = \sqrt{9(x+2)} = \sqrt{9x+18}.$$
But when you try to push a negative number inside a square root, you have to leave its minus sign behind:
$$-3\sqrt{x+2} = -\sqrt{9(x+2)} = -\sqrt{9x+18},$$
because $\sqrt{9} = 3$ not $-3.$
When your number is a variable, it's not as obvious that there is a minus sign.  If $x$ is negative, then $-x$ is positive.  So in your problem the calculation is
$$\frac{1}{x}\sqrt{x^2-9} = -\frac{1}{-x}\sqrt{x^2-9} = -\frac{1}{\sqrt{(-x)^2}}\sqrt{x^2-9} $$
$$= -\frac{1}{\sqrt{x^2}}\sqrt{x^2-9}=-\sqrt{\frac{x^2-9}{x^2}}.$$
A: Maybe you can change the variable
$$y = -x.$$
When $x \rightarrow -\infty$, then $y \rightarrow \infty$.
So that the problem becomes
$$\lim_{y \to \infty} \frac{\sqrt{y^2-9}}{-2y-6}= -\frac{1}{2}.$$
A: The minus sign derives from the absolute value:
\begin{equation}
\dfrac{\sqrt{x^2-9}}{2x-6}\sim\dfrac{\sqrt{x^2}}{2x}=\dfrac{|x|}{2x}=-\dfrac{x}{2x}\to-\dfrac{1}{2} \quad \text{as}\quad x\to-\infty
\end{equation}
