Use the \(\epsilon \)-\(\delta \) definition of a limit to prove the statement \(\lim\limits_{x\rightarrow 3}(4x-5)\neq 10\).

Solution We use a proof by contradiction.* Assume \( \lim\limits_{x\rightarrow 3}(4x-5)=10\) and choose \(\epsilon =1\). (Any smaller positive number \(\epsilon \) will also work.) Then there is a number \(\delta >0\), so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-3\vert \lt\delta\qquad \hbox{then}\quad \vert ( 4x-5) -10\vert \lt1\ \end{equation*}

We simplify the right inequality. \begin{equation*} \begin{array}{rcl} \vert (4x-5)-10 \vert &=& \vert 4x-15 \vert \lt1 \\[3pt] -1&\lt&4x-15\lt 1 \\[3pt] 14&\lt&4x\lt 16 \\[3pt] 3.5&\lt&x\lt 4 \end{array} \end{equation*}

*In a proof by contradiction, we assume that the conclusion is not true and then show this leads to a contradiction.

According to our assumption, whenever \(0\lt\left\vert x-3\right\vert \lt\delta ,\) then \(3.5\lt x\lt 4\). Regardless of the value of \(\delta\), the inequality \( 0\lt \left\vert x-3\right\vert \lt \delta \) is satisfied by a number \(x\) that is less than \(3\). This contradicts the fact that \(3.5\lt x\lt 4\). The contradiction means that \(\lim\limits_{x\rightarrow 3}(4x-5)\neq 10.\)