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

$$\lim_{x\to\infty}(\sqrt{x^2+3x}-\sqrt{x^2-2x})=\frac{5}{2}$$
$$A= \lim_{x\to\infty}(\sqrt{x^2+3x}-\sqrt{x^2-2x})$$ We cannot examine the behavior of $(\sqrt{x^2+3x}-\sqrt{x^2-2x})$ as $x$ approaches $\infty$ right away, because it will lead to the unsolved situation of $\sqrt{\infty}-\sqrt{\infty}=\infty-\infty$, 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: $$A=\lim_{x\to\infty}\Big[(\sqrt{x^2+3x}-\sqrt{x^2-2x})\times\frac{(\sqrt{x^2+3x}+\sqrt{x^2-2x})}{(\sqrt{x^2+3x}+\sqrt{x^2-2x})}\Big]$$ $$A=\lim_{x\to\infty}\frac{(x^2+3x)-(x^2-2x)}{\sqrt{x^2+3x}+\sqrt{x^2-2x}}=\lim_{x\to\infty}\frac{5x}{\sqrt{x^2+3x}+\sqrt{x^2-2x}}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$: $$A=\lim_{x\to\infty}\frac{5}{\frac{\sqrt{x^2+3x}}{x}+\frac{\sqrt{x^2-2x}}{x}}$$ 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, $x = \sqrt{x^2}$. $$A=\lim_{x\to\infty}\frac{5}{\frac{\sqrt{x^2+3x}}{\sqrt{x^2}}+\frac{\sqrt{x^2-2x}}{\sqrt{x^2}}}=\lim_{x\to\infty}\frac{5}{\sqrt{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}}}$$ $$A=\frac{5}{\sqrt{1+0}+\sqrt{1-0}}=\frac{5}{1+1}=\frac{5}{2}$$