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

$$F(x)=\sqrt{9-x}\hspace{1cm}c=5$$ For $F$ to be continuous at $c=5$, $\lim_{x\to 5}F(x)=F(5)=\sqrt{9-5}=\sqrt4=2$ In other words, we need to prove that $\lim_{x\to 5}F(x)=2$. According to the formal definition of limit, for any values of $\epsilon\gt0$, there must exist a corresponding value of $\delta\gt0$ such that for all $x$, $$0\lt|x-5|\lt\delta\Rightarrow|F(x)-2|\lt\epsilon$$ Examining the inequality: $$|F(x)-2|\lt\epsilon$$ $$|\sqrt{9-x}-2|\lt\epsilon$$ $$-\epsilon\lt\sqrt{9-x}-2\lt\epsilon$$ $$2-\epsilon\lt \sqrt{9-x}\lt2+\epsilon$$ $$4-4\epsilon+\epsilon^2\lt9-x\lt4+4\epsilon+\epsilon^2$$ $$-5-4\epsilon+\epsilon^2\lt -x\lt-5+4\epsilon+\epsilon^2$$ $$5+4\epsilon-\frac{\epsilon^2}{2}\gt x\gt5-4\epsilon-\frac{\epsilon^2}{2}$$ So if $x\in(5-4\epsilon-\frac{\epsilon^2}{2},5+4\epsilon-\frac{\epsilon^2}{2})$ then $|F(x)-2|\lt\epsilon$ Take $\delta$ to be the distance from $c=5$ to the nearer endpoint of $(5-4\epsilon-\frac{\epsilon^2}{2},5+4\epsilon-\frac{\epsilon^2}{2})$, or $\delta=\min\Big(5-(5-4\epsilon-\frac{\epsilon^2}{2}),(5+4\epsilon-\frac{\epsilon^2}{2})-5\Big)=\min(4\epsilon+\frac{\epsilon^2}{2},4\epsilon-\frac{\epsilon^2}{2})$. If $\delta$ has this value or any smaller values, then because $0\lt|x-5|\lt\delta$, $x$ will always lie in the interval $(5-4\epsilon-\frac{\epsilon^2}{2},5+4\epsilon-\frac{\epsilon^2}{2})$ so that $|F(x)-2|\lt\epsilon$. Therefore, $\lim_{x\to5}F(x)=F(5)=2$. $F$ is continuous at $x=5$.