Find the limit of the following sequence or determine that the sequence diverges. \( \left\{\frac{n}{\sqrt{49 n^{2}+1}}\right\} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The limit of the sequence is B. The sequence diverges.

To find the limit of the sequence \( \left\{\frac{n}{\sqrt{49 n^{2}+1}}\right\} \), we can start by simplifying the expression. We can factor out \( n^2 \) from the square root in the denominator: \[ \frac{n}{\sqrt{49 n^{2}+1}} = \frac{n}{\sqrt{n^{2}(49 + \frac{1}{n^{2}})}} = \frac{n}{n\sqrt{49 + \frac{1}{n^{2}}}} = \frac{1}{\sqrt{49 + \frac{1}{n^{2}}}} \] As \( n \) approaches infinity, the term \( \frac{1}{n^{2}} \) approaches 0, so we have: \[ \lim_{n \to \infty} \frac{1}{\sqrt{49 + \frac{1}{n^{2}}}} = \frac{1}{\sqrt{49}} = \frac{1}{7} \] Thus, the limit of the sequence is \( \frac{1}{7} \). So, the correct choice is: A. The limit of the sequence is \( \frac{1}{7} \).