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

$$\lim_{x\to\infty}(\sqrt{x^2+25}-\sqrt{x^2-1})=0$$
$$A= \lim_{x\to\infty}(\sqrt{x^2+25}-\sqrt{x^2-1})$$ We cannot examine the behavior of $(\sqrt{x^2+25}-\sqrt{x^2-1})$ as $x$ approaches $\infty$ right away, because it will lead to the unsolved situation of $\infty-\infty$, 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: $$A=\lim_{x\to\infty}\Big[(\sqrt{x^2+25}-\sqrt{x^2-1})\times\frac{(\sqrt{x^2+25}+\sqrt{x^2-1})}{(\sqrt{x^2+25}+\sqrt{x^2-1})}\Big]$$ $$A=\lim_{x\to\infty}\frac{(x^2+25)-(x^2-1)}{(\sqrt{x^2+25}+\sqrt{x^2-1})}=\lim_{x\to\infty}\frac{26}{\sqrt{x^2+25}+\sqrt{x^2-1}}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$: $$A=\lim_{x\to\infty}\frac{\frac{26}{ x}}{\frac{\sqrt{x^2+25}}{x}+\frac{\sqrt{x^2-1}}{x}}=\lim_{x\to\infty}\frac{\frac{26}{x}}{\sqrt{1+\frac{25}{x^2}}+\sqrt{1-\frac{1}{x^2}}}$$ $$A=\frac{0}{\sqrt{1+0}+\sqrt{1-0}}=0$$