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: $$\frac{\sqrt{x^2-9}}{x}=\sqrt{\frac{x^2-9}{x^2}}=\sqrt{1-\frac{9}{x^2}}$$
- Denominator: $$\frac{2x-6}{x}=2-\frac{6}{x}$$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{\sqrt{1-\frac{9}{x^2}}}{2-\frac{6}{x}}$$$$A=\frac{\sqrt{1-9\times0}}{2-6\times0}$$$$A=\frac{1}{2}$$
