## Thomas' Calculus 13th Edition

To prove $\lim\limits_{x \to 9}\sqrt {x-5}=2$, let $c=9, f(x)=\sqrt {x-5}, L=2$. For any given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ such that for any $x, 0\lt|x-c|\lt\delta, |f(x)-L|\lt\epsilon$. This requires $|\sqrt {x-5}-2|\lt\epsilon$, or $-\epsilon\lt \sqrt {x-5}-2\lt\epsilon$ or $2-\epsilon\lt \sqrt {x-5}\lt 2+\epsilon$, which gives $(2-\epsilon)^2-4\lt x-9\lt(2+\epsilon)^2-4$ The above process indicates that if we let $\delta$ be the smaller of $|(2-\epsilon)^2-4|$ and $|(2+\epsilon)^2-4|$, we should have for any $x,0\lt|x-9|\lt\delta$, such that $|f(x)-2|\lt\epsilon$, which proves the limit statement.