Question

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

$$\lim_{x \to \infty} \frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}$$

select the correct choice below and, if necessary, fill in the answer box within your choice.

a. $$\lim_{x \to \infty} \frac{3x^{3}+8x - 5}{8x^{4}-9x^{3}-2}=$$ (simplify your answer.)
b. the limit does not exist and is neither $$\infty$$ nor $$-\infty$$.

Explanation:

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

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

Step2: Use the limit property \(\lim_{x

ightarrow\infty}\frac{1}{x^{n}} = 0\) (\(n>0\))
We know that \(\lim_{x
ightarrow\infty}\frac{3}{x}=0\), \(\lim_{x
ightarrow\infty}\frac{8}{x^{3}} = 0\), \(\lim_{x
ightarrow\infty}\frac{5}{x^{4}}=0\), \(\lim_{x
ightarrow\infty}\frac{9}{x}=0\) and \(\lim_{x
ightarrow\infty}\frac{2}{x^{4}}=0\)

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

Answer:

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