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

$$\lim_{x\to-\infty}\frac{4-3x^3}{\sqrt{x^6+9}}=3$$
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator. Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to-\infty}\frac{4-3x^3}{\sqrt{x^6+9}}$$ The highest power of $x$ in the denominator here, as you can see, is $x^6$. However, as $x^6$ is wrapped inside a square root, to eliminate $x^6$ in square root, we only need to divide both numerator and denominator by $x^3$: $$A=\lim_{x\to-\infty}\frac{\frac{4}{x^3}-3}{\frac{\sqrt{x^6+9}}{x^3}}$$ In the denominator, to put $x^3$ inside the square root, remember that $|x^3|=\sqrt{x^6}$ Here since $x\to-\infty$, we consider only values of $x\lt0$, so $x^3=-\sqrt{x^6}$. $$A=\lim_{x\to-\infty}\frac{\frac{4}{x^3}-3}{\frac{\sqrt{x^6+9}}{-\sqrt{x^6}}}=\lim_{x\to-\infty}\frac{\frac{4}{x^3}-3}{-\sqrt{\frac{x^6+9}{x^6}}}$$ $$A=-\lim_{x\to-\infty}\frac{\frac{4}{x^3}-3}{\sqrt{1+\frac{9}{x^6}}}=-\frac{\lim_{x\to-\infty}\frac{4}{x^3}-\lim_{x\to-\infty}3}{\sqrt{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{9}{x^6}}}$$ $$A=-\frac{0-3}{\sqrt{1+0}}=-\Big(\frac{-3}{\sqrt1}\Big)=3$$