The limit of the sequence $a _ { n } = \frac { - 4 n ^ { 2 } - n - 8 } { 3 n ^ { 2 } + 2 n - 5 }$ is

A)0
B) $- \frac { 4 } { 3 }$
C) $- \frac { 3 } { 4 }$
D) $- \frac { 1 } { 2 }$
E)$\infty$

Question

The limit of the sequence $a _ { n } = \frac { - 4 n ^ { 2 } - n - 8 } { 3 n ^ { 2 } + 2 n - 5 }$ is

A)0
B) $- \frac { 4 } { 3 }$
C) $- \frac { 3 } { 4 }$
D) $- \frac { 1 } { 2 }$
E)$\infty$

Quizplus · Accepted Answer

To find the limit of the sequence as $n$ approaches infinity, divide every term by $n^2$, the highest power of $n$ in the sequence. This simplifies the expression to $\lim_{n \to \infty} \frac{-4 - \frac{1}{n} - \frac{8}{n^2}}{3 + \frac{2}{n} - \frac{5}{n^2}}$. As $n$ approaches infinity, the terms $\frac{1}{n}$ and $\frac{8}{n^2}$ in the numerator and $\frac{2}{n}$ and $\frac{5}{n^2}$ in the denominator approach 0, leaving $\frac{-4}{3}$.

The Limit of the Sequence an=−4n2−n−83n2+2n−5a _ { n } = \frac { - 4 n ^ { 2 } - n - 8 } { 3 n ^ { 2 } + 2 n - 5 }an​=3n2+2n−5−4n2−n−8​

The Limit of the Sequence $a _ { n } = \frac { - 4 n ^ { 2 } - n - 8 } { 3 n ^ { 2 } + 2 n - 5 }$