## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\infty}(\sqrt{9x^2-x}-3x)=-\frac{1}{6}$$
$$A= \lim_{x\to\infty}(\sqrt{9x^2-x}-3x)$$ We cannot examine the behavior of $(\sqrt{9x^2-x}-3x)$ as $x$ approaches $\infty$ right away, because it will lead to the unsolved situation of $\sqrt{\infty}-\infty=\infty-\infty$, which we would try to avoid. Instead, we would want to turn it into a rational function and apply the usual method, by doing the followings: $$A=\lim_{x\to\infty}\Big[(\sqrt{9x^2-x}-3x)\times\frac{(\sqrt{9x^2-x}+3x)}{(\sqrt{9x^2-x}+3x)}\Big]$$ $$A=\lim_{x\to\infty}\frac{9x^2-x-9x^2}{(\sqrt{9x^2-x}+3x)}=\lim_{x\to\infty}\frac{-x}{\sqrt{9x^2-x}+3x}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$: $$A=\lim_{x\to\infty}\frac{-1}{\frac{\sqrt{9x^2-x}}{x}+3}$$ However, we must be careful with the sign here when we try to put $x$ inside the square root: - We know that $|x|=\sqrt{x^2}$. Here, since $x\to\infty$, we consider values of $x\gt0$, therefore, $x = \sqrt{x^2}$. $$A=\lim_{x\to\infty}\frac{-1}{\frac{\sqrt{9x^2-x}}{\sqrt{x^2}}+3}=\lim_{x\to\infty}\frac{-1}{\sqrt{9-\frac{1}{x}}+3}$$ $$A=\frac{-1}{\sqrt{9-0}+3}=\frac{-1}{3+3}=-\frac{1}{6}$$