Determine the limit of the sequence or state that the sequence diverges.

Determine the limit of the sequence or state that the sequence diverges. a_(n)=(n)/(\sqrt(n^(3)+13)) (Use symbolic notation and fractions where needed. Enter DNE if the sequence diverges.) \lim_(n->\infty )a_(n)=1.0 Incorrect Answer

The Correct Answer and Explanation is:

To determine the limit of the sequence an=nn3+13a_n = \frac{n}{\sqrt{n^3 + 13}}an​=n3+13​n​, we need to analyze the behavior of the sequence as n→∞n \to \inftyn→∞.

Step-by-step solution:

1. Express the sequence:
   We are given the sequence: an=nn3+13a_n = \frac{n}{\sqrt{n^3 + 13}}an​=n3+13​n​
2. Factor out the dominant term:
   As nnn grows large, the term n3n^3n3 will dominate over the constant 13 inside the square root. To better understand the limit, factor n3n^3n3 out of the square root: an=nn3(1+13n3)a_n = \frac{n}{\sqrt{n^3 (1 + \frac{13}{n^3})}}an​=n3(1+n313​)​n​ Simplifying further: an=nn3/21+13n3a_n = \frac{n}{n^{3/2} \sqrt{1 + \frac{13}{n^3}}}an​=n3/21+n313​​n​ Now, we can simplify the expression: an=1n1/21+13n3a_n = \frac{1}{n^{1/2} \sqrt{1 + \frac{13}{n^3}}}an​=n1/21+n313​​1​
3. Take the limit as n→∞n \to \inftyn→∞:
   • As n→∞n \to \inftyn→∞, the term 13n3\frac{13}{n^3}n313​ tends to 0. Therefore, the square root term 1+13n3\sqrt{1 + \frac{13}{n^3}}1+n313​​ approaches 1=1\sqrt{1} = 11​=1.
   • Thus, the expression simplifies to:
   an≈1n1/2a_n \approx \frac{1}{n^{1/2}}an​≈n1/21​
   • As n→∞n \to \inftyn→∞, 1n1/2\frac{1}{n^{1/2}}n1/21​ tends to 0.

Conclusion:

Therefore, the limit of the sequence is: lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0n→∞lim​an​=0

This means that the sequence converges to 0 as n→∞n \to \inftyn→∞. The answer is 0.