## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to\infty}\frac{\sqrt{1+4x^6}}{2-x^3}=-2$$
$$A=\lim\limits_{x\to\infty}\frac{\sqrt{1+4x^6}}{2-x^3}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^3$, we have - In the numerator: $X=\frac{\sqrt{1+4x^6}}{x^3}=\frac{\sqrt{1+4x^6}}{\sqrt{x^6}}=\sqrt{\frac{1+4x^6}{x^6}}=\sqrt{\frac{1}{x^6}+4}$ - In the denominator: $Y=\frac{2-x^3}{x^3}=\frac{2}{x^3}-1$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{\sqrt{\frac{1}{x^6}+4}}{\frac{2}{x^3}-1}$$$$A=\frac{\lim\limits_{x\to\infty}(\sqrt{\frac{1}{x^6}+4})}{\lim\limits_{x\to\infty}(\frac{2}{x^3})-1}$$$$A=\frac{\sqrt{\lim\limits_{x\to\infty}(\frac{1}{x^6})+4}}{2\times\lim\limits_{x\to\infty}(\frac{1}{x^3})-1}$$$$A=\frac{\sqrt{0+4}}{2\times0-1}$$$$A=-2$$