University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to-\infty}\Big(\frac{1-x^3}{x^2+7x}\Big)^{5}=\infty$$
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 $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to-\infty}\Big(\frac{1-x^3}{x^2+7x}\Big)^{5}$$ The highest power of $x$ in the denominator here is $x^2$, so we divide both numerator and denominator by $x^2$: $$A=\lim_{x\to-\infty}\Big(\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}\Big)^{5}=\Big(\frac{\lim_{x\to-\infty}\frac{1}{x^2}-\lim_{x\to-\infty}(x)}{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{7}{x}}\Big)^{5}$$ $$A=\Big(\frac{0-\lim_{x\to-\infty}(x)}{1+0}\Big)^5=(-\lim_{x\to-\infty}x)^5=-\lim_{x\to-\infty}x^5$$ As $x$ approaches $-\infty$, $x^5$ will approach $-\infty$. Therefore, $$A=-\lim_{x\to-\infty}x^5=-(-\infty)=\infty$$