## University Calculus: Early Transcendentals (3rd Edition)

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta$ such that for all $x$ $$0\lt|x-3|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
$$\lim_{x\to3}(3x-7)=2$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta$ such that for all $x$ $$0\lt|x-3|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$ 1) Solve the inequality $|f(x)-2|\lt\epsilon$ to find an open interval containing $x=3$ on which the inequality holds for all $x\ne3$ $$|f(x)-2|\lt\epsilon$$ $$|3x-7-2|\lt\epsilon$$ $$|3x-9|\lt\epsilon$$ $$-\epsilon\lt3x-9\lt\epsilon$$ $$-\epsilon+9\lt3x\lt\epsilon+9$$ $$-\frac{\epsilon}{3}+3\lt x\lt\frac{\epsilon}{3}+3$$ So the open interval on which the inequality holds is $(-\frac{\epsilon}{3}+3,\frac{\epsilon}{3}+3)$. 2) Find a value of $\delta\gt0$ that places the centered interval $(3-\delta,3+\delta)$ inside the interval $(-\frac{\epsilon}{3}+3,\frac{\epsilon}{3}+3)$ Take $\delta$ to be the distance from $3$ to the nearer endpoint of $(-\frac{\epsilon}{3}+3,\frac{\epsilon}{3}+3)$. In fact, we can see that the distance from $3$ to either of the two endpoints is the same, which equals $\frac{\epsilon}{3}$. So if $\delta=\frac{\epsilon}{3}$ or any smaller positive value, then the constraint $0\lt|x-3|\lt\delta$ will automatically place $x$ between $-(\epsilon/3)+3$ and $(\epsilon/3)+3$ so that $|f(x)-2|\lt\epsilon$. That means for all $x$, $$0\lt|x-3|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$ This completes our proof.