Answer
$$\lim\limits_{x\to-\infty}\frac{\sqrt{1+4x^6}}{2-x^3}=2$$
Work Step by Step
$$A=\lim\limits_{x\to-\infty}\frac{\sqrt{1+4x^6}}{2-x^3}$$$$A=\lim\limits_{x\to-\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $x^3$, we have
- In the numerator:
Notice that as $x\to-\infty$, we have $x\lt0$.
Therefore, $\sqrt{x^6}=|x^3|=(|x|)^3=(-x)^3=-x^3$
So,
$X=\frac{\sqrt{1+4x^6}}{x^3}=\frac{\sqrt{1+4x^6}}{-\sqrt{x^6}}=-\sqrt{\frac{1+4x^6}{x^6}}=-\sqrt{\frac{1}{x^6}+4}$
- In the denominator:
$Y=\frac{2-x^3}{x^3}=\frac{2}{x^3}-1$
Therefore, $$A=\lim\limits_{x\to-\infty}\frac{-\sqrt{\frac{1}{x^6}+4}}{\frac{2}{x^3}-1}$$$$A=-\frac{\lim\limits_{x\to-\infty}(\sqrt{\frac{1}{x^6}+4})}{\lim\limits_{x\to-\infty}(\frac{2}{x^3})-1}$$$$A=-\frac{\sqrt{\lim\limits_{x\to-\infty}(\frac{1}{x^6})+4}}{2\times\lim\limits_{x\to-\infty}(\frac{1}{x^3})-1}$$$$A=-\frac{\sqrt{0+4}}{2\times0-1}$$$$A=2$$
