Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 14

Determine the values of $\delta$ that correspond to $\varepsilon = 0.1,$ $\varepsilon = 0.05$, and
$\varepsilon = 0.01$ for the $\lim\limits_{x \to 2}(5x-7) = 3$

@ $\varepsilon = 0.1$,


Based from the definition,

$
\begin{equation}
\begin{aligned}

\begin{array}{c}
\text{ if } |x -a| < \delta \text{ then } |f(x) - L| < \varepsilon\\
\text{ if } |x -2| < \delta \text{ then } |(5x-7)-3| < \varepsilon

\end{array}

\end{aligned}
\end{equation}
$


To satisfy inequality $| x - 2 | < \delta $

We want,

$
\begin{equation}
\begin{aligned}
|(5x-7)-3 |& < 0 .1 && \text{(Substitute the given values.)}\\
|5x-7-3 |& < 0 .1 && \text{(Simplify.)}\\
|5x-10 |& < 0 .1 && \text{(Factor.)}\\
5|x-2| & < 0 .1 && \text{(Divde both sides by 5.)}\\
|x-2| & < \displaystyle\frac{0.1}{5} && \text{(Simplify)}\\
|x-2| & < 0.02
\end{aligned}
\end{equation}\\
$


Hence,
$\quad \delta = 0.02$




@ $\varepsilon = 0.05$

$
\begin{equation}
\begin{aligned}
|(5x-7)-3 |& < 0 .05 && \text{(Substitute the given values.)}\\
|5x-7-3 |& < 0 .05 && \text{(Simplify.)}\\
|5x-10 |& < 0 .05 && \text{(Factor.)}\\
5|x-2| & < 0 .05 && \text{(Divde both sides by 5.)}\\
|x-2| & < \displaystyle\frac{0.05}{5} && \text{(Simplify)}\\
|x-2| & < 0.01
\end{aligned}
\end{equation}\\
$

Hence,
$\quad \delta = 0.01$

@ $\varepsilon = 0.01$

$
\begin{equation}
\begin{aligned}
|(5x-7)-3 |& < 0 .01 && \text{(Substitute the given values.)}\\
|5x-7-3 |& < 0 .01 && \text{(Simplify.)}\\
|5x-10 |& < 0 .01 && \text{(Factor.)}\\
5 |x-2| & < 0 .01 && \text{(Divde both sides by 5.)}\\
|x-2| & < \displaystyle\frac{0.01}{5} && \text{(Simplify)}\\
|x-2| & < 0.002
\end{aligned}
\end{equation}\\
$

Hence,
$\quad \delta = 0.002$