Limits of sequences

$L$ is unique as per the Uniqueness of Convergence proposition. The uniqueness allows us to create the limit notation $\underset{n\to \mathrm{\infty }}{lim}{a}_n$, dependent only on the sequence $(a_n{)}_{n=m}^{\mathrm{\infty }}$, 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 $(a_n{)}_{n=m}^{\mathrm{\infty }}$, as the starting index is irrelevant and neglected. So the $\underset{n\to \mathrm{\infty }}{lim}{a}_n$ is the limit for any sequences $(a_n{)}_{n=N}^{\mathrm{\infty }}$ where $N$ is some integer.

Example

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$, proof

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$ is true if the sequence $(a_n{)}_{n=1}^{\mathrm{\infty }}$ converges to 0, where $a_n=\frac{1}{n}$.

Let $\epsilon >0$ be an arbitary real. We must show that there exists an integer $N$ such that $|a_n-0|\le \epsilon$ for all $n\ge N$.

$|a_n-0|=|\frac{1}{n}-0|=\frac{1}{n}\le \frac{1}{N}$.

$\frac{1}{N}\le \epsilon$ is true if $N\ge \frac{1}{\epsilon }$. Such an integer $N$ exists as per the Achimedean principal. Thus $|a_n-0|\le \epsilon$ for all $n\ge N$. Thus, $(a_n{)}_{n=1}^{\mathrm{\infty }}$ is eventually $\epsilon$ close to 0 for arbitary $\epsilon >0$. Thus, $(a_n{)}_{n=1}^{\mathrm{\infty }}$ converges to 0. This allows us to write $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$.

Cauchy sequences of reals → convergence of sequences of reals → uniqueness of convergence → limit, the definition → subsumption of formal limits

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