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 $\lim\limits_{x\rightarrow 3}(4x-5)\neq 10.$