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

$$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\infty$$
To solve these exercises, 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) $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3x^7+5x^2-1}{6x^3-7x+3}$$ The highest power of $x$ in the denominator here is $x^3$, so we divide both numerator and denominator by $x^3$: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3x^4+\frac{5}{x}-\frac{1}{x^3}}{6-\frac{7}{x^2}+\frac{3}{x^3}}$$ $$\lim_{x\to\infty}f(x)=\frac{\lim_{x\to\infty}(3x^4)+0-0}{6-0+0}=\frac{3\lim_{x\to\infty}(x^4)}{6}=\frac{\lim_{x\to\infty}(x^4)}{2}$$ $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\Big(\frac{x^4}{2}\Big)$$ As $x\to\infty$, $(x^4/2)$ approaches $\infty$ as well. Therefore, $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\Big(\frac{x^4}{2}\Big)=\infty$$ (b) $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{3x^7+5x^2-1}{6x^3-7x+3}$$ Again, we divide both numerator and denominator by $x^3$: $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{3x^4+\frac{5}{x}-\frac{1}{x^3}}{6-\frac{7}{x^2}+\frac{3}{x^3}}$$ $$\lim_{x\to-\infty}f(x)=\frac{\lim_{x\to-\infty}(3x^4)+0-0}{6-0+0}=\frac{3\lim_{x\to-\infty}(x^4)}{6}=\frac{\lim_{x\to-\infty}(x^4)}{2}$$ $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\Big(\frac{x^4}{2}\Big)$$ As $x\to-\infty$, $(x^4)$ approaches $\infty$ and $(x^4/2)$ approaches $\infty$ as a result. Therefore, $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\Big(\frac{x^4}{2}\Big)=\infty$$