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\centerline{\Large {\bf MATH 242 Principles of Analysis }}
\vspace{.1in}
\centerline{{\bf Solution to Problem 2m in Section 3.1}}
\vspace{.2in}

{\bf 2m.}  Prove that
$$
\lim_{x \rightarrow 1} \frac{1}{5x-3} \; = \; \frac{1}{2}
$$
using the $\epsilon$-$\delta$ definition.

\vspace{.1in}

{\bf Proof:}  Let $\epsilon > 0$ be given.  We must find a 
$\delta > 0$ such that 
$$
0 < |x-1| < \delta  \; \Longrightarrow \; |f(x) - \frac{1}{2} | < \epsilon.
$$

Simplifying $|f(x)-L|$ gives
$$
\left| \frac{1}{5x-3} - \frac{1}{2} \right| =
\left| \frac{2 - (5x-3)}{2(5x-3)} \right| =
\left| \frac{5 - 5x}{2(5x-3)} \right| =
\frac{5}{2} \cdot \frac{1}{|5x-3|} \cdot |x-1|
$$
In order to make this smaller than $\epsilon$, we
need to bound the term $1/|5x-3|$.  To do this we want the
denominator to be bounded away from 0.  Note that when
$x=3/5$ this term blows up to $+\infty$.  This is a problem!

Suppose that $\delta \leq 1/5$, then $|x-1| < 1/5$ means
$4/5 < x < 6/5$ and thus we have kept $x$ away from the
danger zone around $x=3/5$.  (Choosing
$\delta \leq 1$ is not restrictive enough.  This doesn't
exclude $x=3/5$.)  Then, since $|5x-3|$ is smallest
when $x=4/5$, we have
$$
\frac{4}{5} < x < \frac{6}{5} \; \Longrightarrow \; \frac{1}{|5x-3|} < 1.
$$
Then, letting $\delta = \mbox{min}\{1/5, 2\epsilon/5 \}$, we have
$$
0 < |x-1| < \delta  \; \Longrightarrow \; |f(x) - \frac{1}{2} | =    
\frac{5}{2} \cdot \frac{1}{|5x-3|} \cdot |x-1|  <  
\frac{5}{2} \cdot |x-1| <
\frac{5}{2} \cdot \frac{2}{5 \epsilon} = \epsilon
$$
which completes the proof.


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