## Thomas' Calculus 13th Edition

To prove $\lim\limits_{x \to 3}(3x-7)=2$, we let $c=3, f(x)=3x-7, L=2$ as used in the definition of limits. For any given small value of $\epsilon\gt0$, we need to find a value of $\delta$ so that for all $x, 0\lt |x-c|\lt\delta$, we should have $|f(x)-L|\lt\epsilon$. or $|3x-7-2|\lt\epsilon, |3(x-3)|\lt\epsilon$. Let $\delta=\epsilon/3$, for all $x, 0\lt|x-3|\lt\delta=\epsilon/3$, we have: $|f(x)-L|=|(3x-7)-2|=3|x-3|\lt\epsilon$ which proves the limit statement.