Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 15

Show that the statement $\lim\limits_{x \to 1} (2x+3) =5$ is correct using the $\varepsilon$, $\delta$ definition of limit and illustrate its graph.




Based from the defintion,


$
\begin{equation}
\begin{aligned}

\phantom{x} \text{if } & 0 < |x - a| < \delta
\qquad \text{ then } \qquad
|f(x) - L| < \varepsilon\\

\phantom{x} \text{if } & 0 < |x - 1| < \delta
\qquad \text{ then } \qquad
|(2x+3)-5| < \varepsilon\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & |(2x+3) - 5| = |2x-2| = |2(x-1)| = 2|x-1| \\
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x-1| < \delta \qquad \text{ then } \qquad 2|x-1| < \varepsilon\\
& \text{That is,} \\
& \phantom{x} & \text{ if } 0 < |x-1| < \delta \qquad \text{ then } \qquad |x-1| < \frac{\varepsilon}{2}\\


\end{aligned}
\end{equation}
$


The statement suggests that we should choose $\displaystyle \delta = \frac{\varepsilon}{2}$.

By proving that the assumed value of $\delta$ will fit the definition...



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-1| < \delta \text{ then, }\\
|(2x+3)-5| & = |2x-2| = 2|x-1| < 2 \delta = 2 \left( \frac{\varepsilon}{2} \right) = \varepsilon\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-1| < \delta \qquad \text{ then } \qquad |(2x+3)-5| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 1} (2x+3) = 5


\end{aligned}
\end{equation}
$