Limits of Absolute Value Of Sequence[Theorem and Proof]

The Limits of absolute Values

Today i will be posting about a further theorem that is an order property of the limits of sequences.yesterday i talked about a theorem which is the squeeze theorem.
Today i will be writing about limits of absolute values, and it states:

\subsection*{Limits Of Absolute Value}
Suppose that $\{s_n\}$ is a convergent sequence.then the sequence $\{|s_n|\}$ is also convergent and
$$\begin{equation*} \lim_{n\rightarrow\infty}|s_n|=|\lim_{n\rightarrow\infty}s_n| \end{equation*}$$

Proof: Let $S=\lim_{n\rightarrow\infty}s_n$ and suppose that $\epsilon>0$. choose $N$ so that
$$\begin{equation*} |s_n-S|<\epsilon \end{equation*}$$
$n\geq{N}$. observe that,because of the triangle inequality, this means that
$$\begin{equation*} ||s_n|-|S||\leq{|s_n-S|}<\epsilon \end{equation*}$$
for all $n\geq{N}$. By definition
$$\begin{equation*} \lim_{n\rightarrow\infty}|s_n|=|S| \end{equation*}$$
which proves our arguments.

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