evaluate the following limits. if needed, enter inf for ∞ and -inf for -∞. (a) lim(x→∞) (7x³ - 2x² - 7x)/(3…

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{3}$. For $\lim_{x\rightarrow\pm\infty}\frac{7x^{3}-2x^{2}-7x}{3 - 3x-9x^{3}}$, we get $\lim_{x\rightarrow\pm\infty}\frac{7-\frac{2}{x}-\frac{7}{x^{2}}}{\frac{3}{x^{3}}-\frac{3}{x^{2}}-9}$.

Step2: Evaluate limits of individual terms

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{2}{x}=0$, $\lim_{x\rightarrow\infty}\frac{7}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{3}{x^{3}}=0$, $\lim_{x\rightarrow\infty}\frac{3}{x^{2}} = 0$. So, $\lim_{x\rightarrow\infty}\frac{7-\frac{2}{x}-\frac{7}{x^{2}}}{\frac{3}{x^{3}}-\frac{3}{x^{2}}-9}=\frac{7 - 0-0}{0 - 0-9}=-\frac{7}{9}$. As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{2}{x}=0$, $\lim_{x\rightarrow-\infty}\frac{7}{x^{2}} = 0$, $\lim_{x\rightarrow-\infty}\frac{3}{x^{3}}=0$, $\lim_{x\rightarrow-\infty}\frac{3}{x^{2}} = 0$. So, $\lim_{x\rightarrow-\infty}\frac{7-\frac{2}{x}-\frac{7}{x^{2}}}{\frac{3}{x^{3}}-\frac{3}{x^{2}}-9}=\frac{7 - 0-0}{0 - 0-9}=-\frac{7}{9}$.

Answer:

(a) $-\frac{7}{9}$ (b) $-\frac{7}{9}$