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+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\vskip18pt plus 6pt minus 6pt\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\let\impl\to
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+\def\problem#1{\bigskip\par\penalty-100\item{#1}}
+
+\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