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\noindent {\bf Question}

\noindent A sequence has its $n^{{\rm th}}$ term given by $u_n
=\frac{3n-1}{4n-5}$.  Write the $1^{{\rm st}}$, $5^{{\rm th}}$,
$10^{{\rm th}}$, $100^{{\rm th}}$, $1000^{{\rm th}}$,
$10,000^{{\rm th}}$, and $100,000^{{\rm th}}$ term of the sequence
in decimal form. Make a {\bf guess} as to the limit of this
sequence as $n\rightarrow\infty$. Using the definition of limit,
verify that the guess you've made is correct.

\medskip
\noindent {\bf Answer}
\begin{itemize}
\item $u_1 = \frac{3(1)-1}{4(1)-5} = \frac{2}{-1} = -2$;
\item $u_5 = \frac{3(5)-1}{4(5)-5} = \frac{14}{15} \approx 0.9333$;
\item $u_{10} = \frac{3(10)-1}{4(10)-5} = \frac{29}{35} \approx
0.8286$;
\item $u_{100} = \frac{3(100)-1}{4(100)-5} = \frac{299}{395} \approx
.7570$;
\item $u_{1000} = \frac{3(1000)-1}{4(1000)-5} = \frac{2999}{3995}
\approx 0.7507$;
\item $u_{10000} = \frac{3(10000)-1}{4(10000)-5} = \frac{29999}{39995}
\approx 0.7501$;
\item $u_{100000} = \frac{3(100000)-1}{4(100000)-5} =
\frac{299999}{399995} \approx 0.7500$;
\end{itemize}

\medskip
\noindent So, it seems that a reasonable guess would be that $L
=\lim_{n\rightarrow\infty} u_n$ exists and equals $0.75 =
\frac{3}{4}$.  To verify this, we use the definition: we need to
show that for any choice of $\varepsilon >0$, we can find $M$ so
that $|u_n -L| <\varepsilon$ for all $n >M$.

\medskip
\noindent Calculating, we see that
\[ |u_n -L| = \left| \frac{3n-1}{4n-5} - \frac{3}{4}\right| = \left|
\frac{4(3n-1) - 3(4n-5)}{4(4n-5)} \right| = \left|
\frac{11}{4(4n-5)} \right| = \frac{11}{4(4n-5)}. \] (The last
equality follows since $u_n -L$ is positive for $n >1$.)

\medskip
\noindent To find the value of $M$ so that $|u_n -L| <\varepsilon$
for $n>M$, we start by solving for $n$: since $\frac{11}{4(4n-5)}
<\varepsilon$, we have that  $\frac{11}{4\varepsilon} < 4n-5$, and
so $\frac{11}{16\varepsilon} + \frac{5}{4} < n$.  That is, for a
specified value of $\varepsilon$, we can take $M =
\frac{11}{16\varepsilon} + \frac{5}{4} =
\frac{11+20\varepsilon}{16\varepsilon}$.  Then, for any choice of
$\varepsilon >0$, we set $M =
\frac{11+20\varepsilon}{16\varepsilon}$, and then if we take $n
>M$, working backwards we have that $|u_n -L| <\varepsilon$.

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