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:
[tex] a_n = \frac{5n}{\sqrt{81n^2 +5}}[/tex]
And we can find the limit when x tend to infinity:
[tex] \lim_{n\to\infty} \frac{5n}{\sqrt{81n^2 +5}} [/tex]
And we got this:
[tex] 5 \lim_{n\to\infty} \frac{n}{\sqrt{81n^2 +5}} [/tex]
Now we can divide by n the numerator and denominator like this:
[tex] 5 \lim_{n\to\infty} \frac{1}{\sqrt{81 +\frac{5}{n^2}}} [/tex]
And now we can apply properties of limits and we got this:
[tex] 5 \frac{\lim_{n\to\infty} 1}{\lim_{n\to\infty} \sqrt{81 +\frac{5}{n^2}}} [/tex]
And we got:
[tex] 5 \frac{1}{9}=\frac{5}{9}[/tex]
So then our sequence is bounded by [tex]\frac{5}{9}[/tex]
By definition a monotonic sequence is a "sequence that is always increasing or decreasing".
For this case if we find:
[tex] a_1 = \frac{5}{\sqrt{86}}[/tex]
[tex] a_2 = \frac{10}{\sqrt{329}}[/tex]
We see that [tex] a_2 >a_1[/tex] and in general we see that [tex] a_n >a_{n-1}[/tex]. So then the ebst answer for this case is:
C. The sequence is monotonic and bounded. The limit is 5/9
