Answer
See explanations.
Work Step by Step
Step 1. To prove that the function $F(x)=\sqrt {9-x}$ is continuous at $c=5$, we need to show that the limit of the function exists at this point and is equal to a value of $F(5)=\sqrt {9-5}=2$.
Step 2. For any small positive $\epsilon\gt0$, we need to find a corresponding value $\delta \gt0$ so that for all $x$ in the interval $|x-5|\lt\delta$, we have $|\sqrt {9-x}-2|\lt\epsilon$,
Step 3. The last inequality can be written as $2-\epsilon\lt \sqrt {9-x}\lt\epsilon+2$ or $(2-\epsilon)^2\lt 9-x \lt(\epsilon+2)^2$, which gives $4-(\epsilon+2)^2 \lt x-5 \lt 4-(2-\epsilon)^2$
Step 4 Choose $\delta$ to be the smaller of $|4-(\epsilon+2)^2|$ and $|4-(2-\epsilon)^2|$
Step 5. With the chosen $\delta$ value, working the above steps backwards, we have for all $x$ in the interval $|x-5|\lt\delta$; we have $|\sqrt {9-x}-2|\lt\epsilon$, which proves that $\lim_{x\to5}F(x)=1=F(5)$
Step 6. We conclude that the function is continuous at $c=5$
