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

use the properties of limits to help decide whether the limit exists. if the limit exists, find its value.\n\n$$\\lim_{x \\to \\infty} \\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}$$\n\nselect the correct choice below and, if necessary, fill in the answer box within your choice.\n\na. $$\\lim_{x \\to \\infty} \\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}=$$ (simplify your answer.)\nb. the limit does not exist and is neither $$\\infty$$ nor $$-\\infty$$.

use the properties of limits to help decide whether the limit exists. if the limit exists, find its value.\n\n$$\\lim_{x \\to \\infty} \\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}$$\n\nselect the correct choice below and, if necessary, fill in the answer box within your choice.\n\na. $$\\lim_{x \\to \\infty} \\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}=$$ (simplify your answer.)\nb. the limit does not exist and is neither $$\\infty$$ nor $$-\\infty$$.

Answer

Explanation:

Step1: Divide numerator and denominator by (x^{4})

$$ \begin{align*} \lim_{x\rightarrow\infty}\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}&=\lim_{x\rightarrow\infty}\frac{\frac{3x^{3}}{x^{4}}+\frac{8x}{x^{4}}-\frac{5}{x^{4}}}{\frac{8x^{4}}{x^{4}}-\frac{9x^{3}}{x^{4}}-\frac{2}{x^{4}}}\ &=\lim_{x\rightarrow\infty}\frac{\frac{3}{x}+\frac{8}{x^{3}}-\frac{5}{x^{4}}}{8-\frac{9}{x}-\frac{2}{x^{4}}} \end{align*} $$

Step2: Use the limit property (\lim_{x\rightarrow\infty}\frac{1}{x^{n}} = 0) ((n>0))

We know that (\lim_{x\rightarrow\infty}\frac{3}{x}=0), (\lim_{x\rightarrow\infty}\frac{8}{x^{3}} = 0), (\lim_{x\rightarrow\infty}\frac{5}{x^{4}}=0), (\lim_{x\rightarrow\infty}\frac{9}{x}=0) and (\lim_{x\rightarrow\infty}\frac{2}{x^{4}}=0)

So, (\lim_{x\rightarrow\infty}\frac{\frac{3}{x}+\frac{8}{x^{3}}-\frac{5}{x^{4}}}{8-\frac{9}{x}-\frac{2}{x^{4}}}=\frac{0 + 0-0}{8-0 - 0})

Answer:

A. (\lim_{x\rightarrow\infty}\frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}=0)