Answer
$$\lim\limits_{x\to-\infty}\frac{\sqrt{x^2-9}}{2x-6}=-\frac{1}{2}$$
Work Step by Step
$$A=\lim\limits_{x\to-\infty}\frac{\sqrt{x^2-9}}{2x-6}$$
Divide both numerator and denominator by $x$ (the highest power of $x$ in the denominator), we have:
- Numerator:
Here since $x\to-\infty$, we only consider the values of $x\lt0$.
So, $\sqrt{x^2}=|x|=-x$. Which means $x=-\sqrt{x^2}$
Therefore, $$\frac{\sqrt{x^2-9}}{x}=-\sqrt{\frac{x^2-9}{x^2}}=-\sqrt{1-\frac{9}{x^2}}$$
- Denominator: $$\frac{2x-6}{x}=2-\frac{6}{x}$$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{-\sqrt{1-\frac{9}{x^2}}}{2-\frac{6}{x}}$$$$A=-\frac{\sqrt{1-9\times0}}{2-6\times0}$$$$A=-\frac{1}{2}$$
