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

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta$ such that for all $x$ $$0\lt|x-4|\lt\delta\Rightarrow|f(x)-5|\lt\epsilon$$
$$\lim_{x\to4}(9-x)=5$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta$ such that for all $x$ $$0\lt|x-4|\lt\delta\Rightarrow|f(x)-5|\lt\epsilon$$ 1) Solve the inequality $|f(x)-5|\lt\epsilon$ to find an open interval containing $x=4$ on which the inequality holds for all $x\ne4$ $$|f(x)-5|\lt\epsilon$$ $$|9-x-5|\lt\epsilon$$ $$|4-x|\lt\epsilon$$ $$-\epsilon\lt4-x\lt\epsilon$$ $$-\epsilon-4\lt-x\lt\epsilon-4$$ $$\epsilon+4\gt x\gt4-\epsilon$$ So the open interval on which the inequality holds is $(4-\epsilon,4+\epsilon)$. 2) Find a value of $\delta\gt0$ that places the centered interval $(4-\delta,4+\delta)$ inside the interval $(4-\epsilon,4+\epsilon)$ Take $\delta$ to be the distance from $4$ to the nearer endpoint of $(4-\epsilon,4+\epsilon)$. In other words, $\delta=\min(4-(4-\epsilon),(4+\epsilon)-4)=\min(\epsilon,\epsilon)=\epsilon$. If $\delta=\epsilon$ or any smaller positive value, then the constraint $0\lt|x-4|\lt\epsilon$ will automatically place $x$ between $4-\epsilon$ and $4+\epsilon$ so that $|f(x)-5|\lt\epsilon$. That means for all $x$, $$0\lt|x-4|\lt\delta\Rightarrow|f(x)-5|\lt\epsilon$$ This completes our proof.