Find the limit of the sequence (if it exists). $\alpha _ { n } = \frac { 2 n } { n ^ { 2 } + 7 }$
A) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 0$
B) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 2$
C) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 7$
D) $\lim_ { n \rightarrow \infty } \alpha _ { n } = - 2$
E)does not exist

Question

Find the limit of the sequence (if it exists). $\alpha _ { n } = \frac { 2 n } { n ^ { 2 } + 7 }$ 
A) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 0$
B) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 2$
C) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 7$
D) $\lim_ { n \rightarrow \infty } \alpha _ { n } = - 2$
E)does not exist

Quizplus · Accepted Answer

To find the limit as $n$ approaches infinity, divide numerator and denominator by $n^2$, yielding $\lim_{n \rightarrow \infty} \frac{2/n}{1+7/n^2}$. As $n$ approaches infinity, $2/n$ and $7/n^2$ approach 0, making the limit $0$.

Find the Limit of the Sequence (If It Exists) αn=2nn2+7\alpha _ { n } = \frac { 2 n } { n ^ { 2 } + 7 }αn​=n2+72n​

Find the Limit of the Sequence (If It Exists) $\alpha _ { n } = \frac { 2 n } { n ^ { 2 } + 7 }$