## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to\infty}(\sqrt{x^2+ax}-\sqrt{x^2+bx})=\frac{a-b}{2}$$
$A=\lim\limits_{x\to\infty}(\sqrt{x^2+ax}-\sqrt{x^2+bx})$ $A=\lim\limits_{x\to\infty}\Bigg[(\sqrt{x^2+ax}-\sqrt{x+bx})\times\frac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\Bigg]$ $A=\lim\limits_{x\to\infty}\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$ $A=\lim\limits_{x\to\infty}\frac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$ $A=\lim\limits_{x\to\infty}\frac{X}{Y}$ Divide both numerator and denominator by $x$, we have - The numerator: $X=\frac{ax-bx}{x}=\frac{(a-b)x}{x}=a-b$ - The denominator: $Y=\frac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{x}=\frac{\sqrt{x^2+ax}}{x}+\frac{\sqrt{x^2+bx}}{x}=\sqrt{\frac{x^2+ax}{x^2}}+\sqrt{\frac{x^2+bx}{x^2}}=\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}$ Therefore, $A=\lim\limits_{x\to\infty}\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$ $A=\frac{\lim\limits_{x\to\infty}(a-b)}{\sqrt{1+\lim\limits_{x\to\infty}(\frac{a}{x})}+\sqrt{1+\lim\limits_{x\to\infty}(\frac{b}{x})}}$ $A=\frac{a-b}{\sqrt{1+a\times0}+\sqrt{1+b\times0}}$ $A=\frac{a-b}{2}$