find the limit, if it exists.\n\n lim _ { x \rightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 }…

find the limit, if it exists.\n\n lim _ { x \rightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 } \n\nselect the correct answer below and, if necessary, fill in the answer box to complete your choice.\n\na. ( lim _ { x \rightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 } = ) (simplify your answer.)\n\nb. the limit does not exist and is neither ( - infty ) nor ( + infty )

find the limit, if it exists.\n\n lim _ { x \rightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 } \n\nselect the correct answer below and, if necessary, fill in the answer box to complete your choice.\n\na. ( lim _ { x \rightarrow infty } \frac { - 4 x ^ { 4 } + 9 x } { 3 x ^ { 3 } - 4 x - 9 } = ) (simplify your answer.)\n\nb. the limit does not exist and is neither ( - infty ) nor ( + infty )

Answer

Explanation:

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

$$\lim_{x\rightarrow\infty}\frac{-4x^{4}+9x}{3x^{3}-4x - 9}=\lim_{x\rightarrow\infty}\frac{\frac{-4x^{4}}{x^{3}}+\frac{9x}{x^{3}}}{\frac{3x^{3}}{x^{3}}-\frac{4x}{x^{3}}-\frac{9}{x^{3}}}$$ Simplify to get $$\lim_{x\rightarrow\infty}\frac{-4x+\frac{9}{x^{2}}}{3-\frac{4}{x^{2}}-\frac{9}{x^{3}}}$$

Step2: Evaluate the limit

As (x\rightarrow\infty), (\frac{9}{x^{2}}\rightarrow0), (\frac{4}{x^{2}}\rightarrow0), (\frac{9}{x^{3}}\rightarrow0) So the limit becomes $$\frac{\lim_{x\rightarrow\infty}(-4x)+\lim_{x\rightarrow\infty}\frac{9}{x^{2}}}{\lim_{x\rightarrow\infty}3-\lim_{x\rightarrow\infty}\frac{4}{x^{2}}-\lim_{x\rightarrow\infty}\frac{9}{x^{3}}}$$ Since (\lim_{x\rightarrow\infty}(-4x)=-\infty) and the denominator approaches (3)

Answer:

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