use the properties of limits to help decide whether the limit exists. if the limit exists, find its value…

Explanation:

Step1: Divide numerator and denominator by highest - power of x in denominator

Divide both the numerator $7x^{3}-3x - 5$ and the denominator $3x^{2}-8x - 7$ by $x^{2}$. We get $\lim_{x\rightarrow\infty}\frac{\frac{7x^{3}}{x^{2}}-\frac{3x}{x^{2}}-\frac{5}{x^{2}}}{\frac{3x^{2}}{x^{2}}-\frac{8x}{x^{2}}-\frac{7}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{7x-\frac{3}{x}-\frac{5}{x^{2}}}{3-\frac{8}{x}-\frac{7}{x^{2}}}$.

Step2: Evaluate the limit of each term

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{3}{x}=0$, $\lim_{x\rightarrow\infty}\frac{5}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{8}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{7}{x^{2}}=0$. Then $\lim_{x\rightarrow\infty}\frac{7x-\frac{3}{x}-\frac{5}{x^{2}}}{3-\frac{8}{x}-\frac{7}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{7x-0 - 0}{3-0 - 0}=\lim_{x\rightarrow\infty}\frac{7x}{3}$.

Step3: Determine the limit value

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{7x}{3}=\infty$. So the limit does not exist.

Answer:

B. The limit does not exist and is neither $\infty$ nor $-\infty$.