Question

Find the limit of the following sequence or determine that the limit does not exist. {5n/Squareroot 81 n^2 + 5} Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The sequence is monotonic, but it is unbounded. The limit is_________---
B. The sequence is not monotonic, but it is bounded. The limit is_____________
C. The sequence is monotonic and bounded. The limit is____________
D. The sequence is monotonic, unbounded, and the limit does not exist.

Answer 1

Answer:

C. The sequence is monotonic and bounded. The limit is 5/9

Step-by-step explanation:

For this case we have the following n term:

$a_n = \frac{5n}{\sqrt{81n^2 +5}}$

And we can find the limit when x tend to infinity:

$\lim_{n\to\infty} \frac{5n}{\sqrt{81n^2 +5}}$

And we got this:

$5 \lim_{n\to\infty} \frac{n}{\sqrt{81n^2 +5}}$

Now we can divide by n the numerator and denominator like this:

$5 \lim_{n\to\infty} \frac{1}{\sqrt{81 +\frac{5}{n^2}}}$

And now we can apply properties of limits and we got this:

$5 \frac{\lim_{n\to\infty} 1}{\lim_{n\to\infty} \sqrt{81 +\frac{5}{n^2}}}$

And we got:

$5 \frac{1}{9}=\frac{5}{9}$

So then our sequence is bounded by $\frac{5}{9}$

By definition a monotonic sequence is a "sequence that is always increasing or decreasing".

For this case if we find:

$a_1 = \frac{5}{\sqrt{86}}$

$a_2 = \frac{10}{\sqrt{329}}$

We see that $a_2 >a_1$ and in general we see that $a_n >a_{n-1}$. So then the ebst answer for this case is:

C. The sequence is monotonic and bounded. The limit is 5/9