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

$$\lim_{x\to-\infty}(\sqrt{x^2+3}+x)=0$$
$$A= \lim_{x\to-\infty}(\sqrt{x^2+3}+x)$$ We cannot examine the behavior of $(\sqrt{x^2+3}+x)$ 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: $$A=\lim_{x\to-\infty}\Big[(\sqrt{x^2+3}+x)\times\frac{(\sqrt{x^2+3}-x)}{(\sqrt{x^2+3}-x)}\Big]$$ $$A=\lim_{x\to-\infty}\frac{(x^2+3)-x^2}{(\sqrt{x^2+3}-x)}=\lim_{x\to-\infty}\frac{3}{\sqrt{x^2+3}-x}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$: $$A=\lim_{x\to-\infty}\frac{\frac{3}{x}}{\frac{\sqrt{x^2+3}}{x}-1}$$ 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 $x\lt0$, therefore, $x = -\sqrt{x^2}$. $$A=\lim_{x\to-\infty}\frac{\frac{3}{x}}{\frac{\sqrt{x^2+3}}{-\sqrt{x^2}}-1}=\lim_{x\to-\infty}\frac{\frac{3}{x}}{-\sqrt{1+\frac{3}{x^2}}-1}$$ $$A=\frac{0}{-\sqrt{1+0}-1}=\frac{0}{-2}=0$$