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\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
2022-04-26 at 11:59 pm}\hrule height .5pt}}

\centerline{\bigbf Homework 12 - Holden Rohrer}
\bigskip

\noindent{\bf Collaborators:} None

\section{Ross Chapter 2, 8.1d, 8.2c, 8.7b}

\problem{8.1d} Prove that $\lim {n+6\over n^2-6} = 0.$

\answer
Let $\epsilon > 0.$
We will show $N$ such that for all $n>N,$ $|{n+6\over n^2-6}-0| <
\epsilon.$

Where $n>6,$
$$|{n+6\over n^2-6}| = {n+6\over n^2-6},$$
and $n^2-6>n^2-36,$ so
$${n+6\over n^2-6} < {n+6\over n^2-36} = {1\over n-6}.$$
Choosing $n > N = 6+1/\epsilon,$ and since $n-6 > N-6,$
$${1\over n-6} < {1\over N-6} = {1\over 6+1/\epsilon-6} = {1\over
1/\epsilon} = \epsilon,$$
showing that $\lim {n+6\over n^2-6} = 0.$
\endanswer

\problem{8.2c} Find the limit of $c_n = {4n+3\over 7n-5}$ and then prove
your claim.

\answer

$c_n$ converges to $4/7.$

Let $\epsilon > 0.$
For all
$$n > N = {41\over 49\epsilon} + {5\over 7},$$
we can show that $|c_n-4/7| < \epsilon.$
$$|c_n-4/7| = \left|{4n+3\over 7n-5}-{4\over 7}\right| = \left|{28n+21 -
(28n-20)\over 49n-35}\right| = \left|{41\over 49n-35}\right|,$$
and from $n > N,$ (and where $N\geq 1$) $49n-35 > 49N-35 > 0,$ so
$$\left|{41\over 49n-35}\right| > {41\over 49N-35} =
{41\over 49({41\over 49\epsilon}+{5\over 7})-35} =
{41\over {41\over\epsilon} + 35-35} =
{41\epsilon\over 41} = \epsilon.$$
We have now shown that $\lim c_n = 4/7.$

\endanswer

\problem{8.7b} Show that $s_n = (-1)^n n$ does not converge.

\answer
{\bf Disproof.}

Assume for the sake of contradiction that $s_n$ converges to $s.$
Let $\epsilon = 1.$
We must have $N$ such that for all $n>N,$ $|s_n-s| < 1$ (implying also
$|s_{n+1}-s| < 1$).
However,
$$|s_n-s_{n+1}| = |s_n-s+s-s_{n+1}| \leq |s_n-s| + |s_{n+1}-s| < 1 + 1 =
2.$$
Yet, $|s_{n+1}-s_n| = |(-1)^n(-(n+1)-n)| = 2n+1,$
and where $n>1,$ we obtain $2n+1>3$ and from the earlier inequality
$2n+1<2,$ completing a contradiction.
We have then shown that $s_n$ does not converge.
\endanswer

\bye