Question

find the limit, if it exists.

lim _ { x
ightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 }

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

a. ( lim _ { x
ightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 } = ) (simplify your answer.)

b. the limit does not exist and is neither ( - infty ) nor ( + infty )

Explanation:

Step1: Divide numerator and denominator by \(x^{3}\)

$$\lim_{x ightarrow\infty}\frac{-4x^{4}+9x}{3x^{3}-4x - 9}=\lim_{x ightarrow\infty}\frac{\frac{-4x^{4}}{x^{3}}+\frac{9x}{x^{3}}}{\frac{3x^{3}}{x^{3}}-\frac{4x}{x^{3}}-\frac{9}{x^{3}}}$$

Simplify to get

$$\lim_{x ightarrow\infty}\frac{-4x+\frac{9}{x^{2}}}{3-\frac{4}{x^{2}}-\frac{9}{x^{3}}}$$

Step2: Evaluate the limit

As \(x
ightarrow\infty\), \(\frac{9}{x^{2}}
ightarrow0\), \(\frac{4}{x^{2}}
ightarrow0\), \(\frac{9}{x^{3}}
ightarrow0\)
So the limit becomes

$$\frac{\lim_{x ightarrow\infty}(-4x)+\lim_{x ightarrow\infty}\frac{9}{x^{2}}}{\lim_{x ightarrow\infty}3-\lim_{x ightarrow\infty}\frac{4}{x^{2}}-\lim_{x ightarrow\infty}\frac{9}{x^{3}}}$$

Since \(\lim_{x
ightarrow\infty}(-4x)=-\infty\) and the denominator approaches \(3\)

Answer:

B. The limit does not exist and is neither \(-\infty\) nor \(+\infty\)