Use the formal definition of the limit of a sequence to prove the following limit. \( \lim _{n \rightarrow \infty}\left(\frac{1}{n}\right)=0 \) Complete the steps of the proof below. To show that \( \lim _{n \rightarrow \infty}\left(\frac{1}{n}\right)=0 \), for every \( \varepsilon>0 \), show that there exists an integer \( N \) such that for all if \( n>N \), then \( |\square-0|<\varepsilon \).

To show that \( \lim _{n \rightarrow \infty}\left(\frac{1}{n}\right)=0 \), for every \( \varepsilon>0 \), we need to find an integer \( N \) such that for all \( n>N \), it follows that \( |\frac{1}{n}-0|<\varepsilon \). We start with the inequality: \[ |\frac{1}{n}-0| = \frac{1}{n} < \varepsilon. \] To solve for \( n \), we can rearrange this inequality: \[ \frac{1}{n} < \varepsilon \implies n > \frac{1}{\varepsilon}. \] Now, we can choose \( N = \lceil \frac{1}{\varepsilon} \rceil \), the smallest integer greater than or equal to \( \frac{1}{\varepsilon} \). Therefore, for all integers \( n > N \): \[ n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon. \] This shows that \( |\frac{1}{n}-0|<\varepsilon \) holds, thus proving: \[ \lim _{n \rightarrow \infty}\left(\frac{1}{n}\right)=0. \] This completes our proof! In conclusion, we have shown that for any positive \( \varepsilon \), there exists an integer \( N \) such that for all \( n > N \), the condition holds true, satisfying the formal definition of the limit of a sequence.