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

$$h(x)=\sqrt{2x-3}\hspace{1cm}c=2$$ For $h$ to be continuous at $c=2$, $\lim_{x\to 2}h(x)=h(2)=\sqrt{2\times2-3}=\sqrt1=1$ In other words, we need to prove that $\lim_{x\to 2}h(x)=1$. 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 $x$, $$0\lt|x-2|\lt\delta\Rightarrow|h(x)-1|\lt\epsilon$$ Examining the inequality: $$|h(x)-1|\lt\epsilon$$ $$|\sqrt{2x-3}-1|\lt\epsilon$$ $$-\epsilon\lt\sqrt{2x-3}-1\lt\epsilon$$ $$1-\epsilon\lt \sqrt{2x-3}\lt1+\epsilon$$ $$1-2\epsilon+\epsilon^2\lt2x-3\lt1+2\epsilon+\epsilon^2$$ $$4-2\epsilon+\epsilon^2\lt2x\lt4+2\epsilon+\epsilon^2$$ $$2-\epsilon+\frac{\epsilon^2}{2}\lt x\lt2+\epsilon+\frac{\epsilon^2}{2}$$ So if $x\in(2-\epsilon+\frac{\epsilon^2}{2},2+\epsilon+\frac{\epsilon^2}{2})$ then $|h(x)-1|\lt\epsilon$ Take $\delta$ to be the distance from $c=2$ to the nearer endpoint of $(2-\epsilon+\frac{\epsilon^2}{2},2+\epsilon+\frac{\epsilon^2}{2})$, or $\delta=\min\Big((2-(2-\epsilon+\frac{\epsilon^2}{2}),(2+\epsilon+\frac{\epsilon^2}{2})-2\Big)=\min(\epsilon-\frac{\epsilon^2}{2},\epsilon+\frac{\epsilon^2}{2})$. If $\delta$ has this value or any smaller values, then because $0\lt|x-2|\lt\delta$, $x$ will always lie in the interval $(2-\epsilon+\frac{\epsilon^2}{2},2+\epsilon+\frac{\epsilon^2}{2})$ so that $|h(x)-1|\lt\epsilon$. Therefore, $\lim_{x\to2}h(x)=h(2)=1$. $h$ is continuous at $x=2$.