Limits of sequences
If a sequence of reals \( (a_n)_{n=m}^{\infty} \) converges to a real \( L \), we say that \( (a_n)_{n=m}^{\infty} \) is convergent and that \( L \) is the limit of the sequence. We write:\( L \) is unique as per the Uniqueness of Convergence proposition. The uniqueness allows us to create the limit notation \( \lim_{n\to\infty} a_n \), dependent only on the sequence \( (a_n)_{n=m}^{\infty} \), and for that notation to be equated to a single real. If fact, the limit notation doesn't depend on the full specification of the sequence \( (a_n)_{n=m}^{\infty} \), as the starting index is irrelevant and neglected. So the \( \lim_{n\to\infty} a_n \) is the limit for any sequences \( (a_n)_{n=N}^{\infty} \) where \( N \) is some integer.
Example
\( \lim_{n\to\infty} \frac{1}{n} = 0 \), proof
\( \lim_{n\to\infty} \frac{1}{n}= 0 \) is true if the sequence \( (a_n)_{n=1}^{\infty} \) converges to 0, where \( a_n = \frac{1}{n} \).
Let \( \varepsilon > 0 \) be an arbitary real. We must show that there exists an integer \( N \) such that \( |a_n - 0| \le \varepsilon \) for all \( n \ge N \).
\( |a_n - 0| = |\frac{1}{n} - 0| = \frac{1}{n} \le \frac{1}{N} \).
\( \frac{1}{N} \le \varepsilon \) is true if \( N \ge \frac{1}{\varepsilon} \). Such an integer \( N \) exists as per the Achimedean principal. Thus \( |a_n - 0| \le \varepsilon \) for all \( n \ge N \). Thus, \( (a_n)_{n=1}^{\infty} \) is eventually \( \varepsilon \) close to 0 for arbitary \( \varepsilon > 0 \). Thus, \( (a_n)_{n=1}^{\infty} \) converges to 0. This allows us to write \( \lim_{n\to\infty} \frac{1}{n}= 0 \).
Cauchy sequences of reals → convergence of sequences of reals → uniqueness of convergence → limit, the definition → subsumption of formal limits
Source
Tao, Analysis IChapter 6