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\noindent {\bf Question}

\noindent Using the definition of limit, prove that
$\lim_{n\rightarrow\infty} \frac{1+2\cdot 10^n}{5+3\cdot 10^n}
=\frac{2}{3}$.  For what value of $M$ do we have that $|
\frac{1+2\cdot 10^n}{5+3\cdot 10^n} -\frac{2}{3} | < 10^{-3}$ for
all $n>M$?

\medskip

\noindent {\bf Answer}

\noindent Set $a_n = \frac{1+2\cdot 10^n}{5+3\cdot 10^n}$ and $L =
\frac{2}{3}$.  For each choice of $\varepsilon
>0$, we need to show that there exists $M$ so that $|a_n -L|
<\varepsilon$ for all $n>M$.

\medskip
\noindent Calculating, we see that
\[ |a_n -L| = \left| \frac{1+2\cdot 10^n}{5+3\cdot 10^n} -\frac{2}{3}
\right| = \left| \frac{3 + 6\cdot 10^n - (10 + 6\cdot 10^n)}{3(5
+3\cdot 10^n)} \right| = \left| \frac{7}{15 +9\cdot 10^n}\right|.
\]

\noindent Hence, for a given value of $\varepsilon >0$, we want to
find $M$ so that $\left| \frac{7}{15 + 9\cdot 10^n} \right|
<\varepsilon$ for $n>M$.  So, we solve for $n$ in terms of
$\varepsilon$.  First, note that $\frac{7}{15 + 9\cdot 10^n} >0$
for all positive integers $n$. So, we need only solve $\frac{7}{15
+ 9\cdot 10^n} <\varepsilon$ for $n$.

\medskip
\noindent So, $\frac{7}{\varepsilon} < 15 + 9\cdot 10^n$, and so
$-15 + \frac{7}{\varepsilon} < 9\cdot 10^n$, and so $-\frac{15}{9}
+ \frac{7}{9\varepsilon} < 10^n$.  Performing a final bit of
simplification, we get $\frac{-15\varepsilon + 7}{9\varepsilon} <
10^n$.  If the numerator is positive, that is if $\varepsilon <
\frac{7}{15}$, we can solve for $n$ by taking $\log_{10}$ of both
sides.  If on the other hand the numerator is negative, then any
positive integer will do.  So, set
\[ M = \left\{ \begin{array}{ll}
    1 & \mbox{ if $\varepsilon \ge \frac{7}{15}$}; \\
    \log_{10}\left( \frac{-15\varepsilon + 7}{9\varepsilon} \right) &
    \mbox{ otherwise} \end{array}\right. \]

\medskip
\noindent To get a specific value of $M$ so that $|a_n
-L|<10^{-3}$ for $n >M$, we substitute $\varepsilon = 10^{-3}$
into the above equation to get that $n > \log_{10} \left(
\frac{-15\cdot 10^{-3} + 7}{9\cdot 10^{-3}} \right) \approx
2.8899$.  So, we can take $M =3$.


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