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.

Important:Do you know how I make money online? It just got easier, you don't have to be a blogger anymore to go through the stress of writing articles, all you need to do is register with paydotcom and become an affiliate marketer, you have the chances of making a thousand dollar every month and more, sign up  for free here..