Find the limit of the sequence if it converges; otherwise indicate divergence.
-$$a _ { n } = \frac { 8 + 5 n + 5 n ^ { 4 } } { 7 n ^ { 4 } - 1 n ^ { 3 } - 7 }$$
A) $- \frac { 5 } { 7 }$
B) $\frac { 8 } { 7 }$
C) $\frac { 5 } { 7 }$
D) Diverges

Question

Quizplus · Accepted Answer

The highest power of $n$ is in the denominator, so we divide both the numerator and denominator by $n^4$ to get
$$a_n = \frac{5/n^3 + 5 + 8/n^4}{7-1/n-7/n^4}.$$
Since the limit of a sum is the sum of the limits (if they exist), we just need to find the limits of each term in the numerator and denominator. We see that
$$\lim_{n	o\infty}\frac{5}{n^3}=0,$$
$$\lim_{n	o\infty}\frac{8}{n^4}=0,$$
$$\lim_{n	o\infty}\frac{1}{n}=0,$$
so the limiting behavior of $a_n$ is determined only by 
$$\lim_{n	o\infty} \frac{5+5}{7} = \frac{5}{7}.$$ 
Thus, the answer is $\boxed{	extbf{(C)}}.$

Find the Limit of the Sequence If It Converges; Otherwise an=8+5n+5n47n4−1n3−7a _ { n } = \frac { 8 + 5 n + 5 n ^ { 4 } } { 7 n ^ { 4 } - 1 n ^ { 3 } - 7 }an​=7n4−1n3−78+5n+5n4​

Find the Limit of the Sequence If It Converges; Otherwise $a _ { n } = \frac { 8 + 5 n + 5 n ^ { 4 } } { 7 n ^ { 4 } - 1 n ^ { 3 } - 7 }$