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+13n, we need to analyze the behavior of the sequence as n→∞n \to \inftyn→∞.
Step-by-step solution:
- Express the sequence:
We are given the sequence: an=nn3+13a_n = \frac{n}{\sqrt{n^3 + 13}}an=n3+13n - 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+n313n 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+n3131 - 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:
- 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: limn→∞an=0\lim_{n \to \infty} a_n = 0n→∞liman=0
This means that the sequence converges to 0 as n→∞n \to \inftyn→∞. The answer is 0.
