Answer
To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-9|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
Work Step by Step
$$\lim_{x\to9}\sqrt{x-5}=2$$
Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-9|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
1) Solve the inequality $|f(x)-2|\lt\epsilon$ to find an open interval containing $x=9$ on which the inequality holds for all $x\ne9$
$$|f(x)-2|\lt\epsilon$$ $$|\sqrt{x-5}-2|\lt\epsilon$$ $$-\epsilon\lt\sqrt{x-5}-2\lt\epsilon$$ $$2-\epsilon\lt\sqrt{x-5}\lt2+\epsilon$$
Square all sides: $$(2-\epsilon)^2\lt x-5\lt(2+\epsilon)^2$$ $$4-4\epsilon+\epsilon^2\lt x-5\lt4+4\epsilon+\epsilon^2$$ $$9-4\epsilon+\epsilon^2\lt x\lt 9+4\epsilon+\epsilon^2$$
So the open interval on which the inequality holds is $( 9-4\epsilon+\epsilon^2, 9+4\epsilon+\epsilon^2)$.
2) Find a value of $\delta\gt0$ that places the centered interval $(9-\delta,9+\delta)$ inside the interval $( 9-4\epsilon+\epsilon^2, 9+4\epsilon+\epsilon^2)$
Take $\delta$ to be the distance from $9$ to the nearer endpoint of $( 9-4\epsilon+\epsilon^2, 9+4\epsilon+\epsilon^2)$. In other words, $\delta=\min(9-(9-4\epsilon+\epsilon^2),(9+4\epsilon+\epsilon^2)-9)=\min(4\epsilon-\epsilon^2,4\epsilon+\epsilon^2)$
So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-9|\lt\delta$ will automatically place $x$ between $9-4\epsilon+\epsilon^2$ and $9+4\epsilon+\epsilon^2$ so that $|f(x)-2|\lt\epsilon$.
That means for all $x$, $$0\lt|x-9|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
This completes our proof.
