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\noindent {\bf Question}

\noindent Prove that each of the following statements is true,
using the definition of limit.
\begin{enumerate}
\item if $x_n\rightarrow -4$ as $n\rightarrow\infty$, then
$\sqrt{|x_n|}\rightarrow 2$ as $n\rightarrow\infty$;
\item if $x_n\rightarrow -4$ as $n\rightarrow\infty$, then
$x_n^2\rightarrow 16$ as $n\rightarrow\infty$;
\item if $x_n\rightarrow -4$ as $n\rightarrow\infty$, then
$\frac{x_n}{3}\rightarrow -\frac{4}{3}$ as $n\rightarrow\infty$;
\end{enumerate}

\medskip

\noindent {\bf Answer}

\noindent In all three of these statements, we start with the same
piece of information, namely that $\lim_{n\rightarrow\infty} x_n
=-4$. That is, for each $\varepsilon
&gt;0$, there exists $M$ (which depends on $\varepsilon$) so that $|x_n -
(-4)| =|x_n +4| &lt;\varepsilon$ for $n &gt;M$.
\begin{enumerate}
\item we need to show that $\lim_{n\rightarrow\infty} \sqrt{|x_n|} =
2$, which is phrased mathematically as needing to show that for
each $\mu &gt;0$, there exists $P$ so that $| \sqrt{|x_n|} -2| &lt; \mu$
for $n
&gt;P$.  We start by rewriting $| \sqrt{|x_n|} -2|$, using the standard
trick for handling differences of square roots, namely
\[ | \sqrt{|x_n|} -2| =|\sqrt{|x_n|} -2|\cdot \frac{|\sqrt{|x_n|}
+ 2|}{|\sqrt{|x_n|} + 2| } = \frac{|\: |x_n| -4|}{|\sqrt{|x_n|} +
2|}\le \frac{|\: |x_n| -4|}{2}. \] (The last inequality follows
from the fact that $|\: \sqrt{|x_n|} +2|\ge 2$ for all possible
values of $x_n$.)  Since for any $\mu &gt;0$, there exists $M$ so
that $ |\: |x_n| -4| &lt; 2\mu$ (by using the definition of
$\lim_{n\rightarrow\infty} |x_n| =4$) for $n &gt;M$, we have that
\[ | \sqrt{|x_n|} -2| \le \frac{|\: |x_n| -4|}{2} &lt; \frac{2\mu}{2}
=\mu \] for $n &gt;M$, and so we are done.
\item we need to show that $\lim_{n\rightarrow\infty} x_n^2 =16$,
which is phrased mathematically as needing to show that for each
$\mu
&gt;0$, there exists $P$ so that $| x_n^2 -16 | &lt; \mu$ for $n &gt;P$.  We
start by rewriting $| x_n^2 - 16|$, using that it is the
difference of two squares:
\[ | x_n^2 - 16| =| (x_n -4)(x_n +4)| =| x_n -4|\: |x_n +4|. \]
Now apply the definition of $\lim_{n\rightarrow\infty} x_n =-4$
with $\varepsilon =1$, so that there exists $M$ so that if $n &gt;N$,
then $|x_n - (-4)| &lt; 1$.  In particular, if $n &gt;M$, then $-5 &lt; x_n
&lt; -3$, and so $|x_n| &lt; 5$, and so $|x_n -4| \le |x_n| + 4 &lt; 9$.

\medskip
\noindent Since $x_n\rightarrow -4$ by assumption, we know that
for any $\varepsilon &gt;0$, there is $Q$ so that $|x_n - (-4)| =
|x_n +4| &lt;\frac{1}{9} \varepsilon$ for $n&gt;Q$.  Hence, if $n &gt; P =
{\rm max}(M,Q)$, then
\[ | x_n^2 -16 | =| x_n -4|\: |x_n +4| &lt; 9 \: \frac{1}{9} \varepsilon
=\varepsilon, \] as desired.
\item we need to show that $\lim_{n\rightarrow\infty} \frac{x_n}{3}
=-\frac{4}{3}$, which is phrased mathematically as needing to show
that for each $\mu &gt;0$, there exists $P$ so that $| \frac{x_n}{3}-
(-\frac{4}{3})| =| \frac{x_n}{3} + \frac{4}{3} | &lt;\mu$ for $n &gt;P$.
Note that $| \frac{x_n}{3}- (-\frac{4}{3} )| =| \frac{x_n}{3}
+\frac{4}{3} =\frac{1}{3}|x_n +4|$.  We know from the definition
of $\lim_{n\rightarrow\infty} x_n =-4$ given above that for any
$\mu &gt;0$, there exists $M$ so that $|x_n - (-4)| =|x_n +4| &lt;3 \mu$
for $n &gt;M$. Hence, for $n &gt;M$, we have that $\frac{1}{3} |x_n +4|
&lt;\frac{1}{3} 3 \mu =\mu$ for $n &gt;M$, and so we are done.
\end{enumerate}

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