Answer
See explanations.
Work Step by Step
Step 1. To prove that the function $h(x)=\sqrt {2x-3}$ is continuous at $c=2$, we need to show that the limit of the function exists at this point and is equal to a value of $h(2)=\sqrt {2(2)-3}=1$.
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 $|x-2|\lt\delta$, we have $|\sqrt {2x-3}-1|\lt\epsilon$,
Step 3. The last inequality can be written as $1-\epsilon\lt \sqrt {2x-3}\lt\epsilon+1$ or $3+(1-\epsilon)^2\lt 2x \lt(\epsilon+1)^2+3$ which gives $\frac{(1-\epsilon)^2-1}{2} \lt x-2 \lt \frac{(\epsilon+1)^2-1}{2}$
Step 4 Choose $\delta$ to be the smaller of $|\frac{(1-\epsilon)^2-1}{2}|$ and $|\frac{(\epsilon+1)^2-1}{2}|$
Step 5. With the chosen $\delta$ value, working the above steps backwards, we have for all $x$ in the interval $|x-2|\lt\delta$, we have $|\sqrt {2x-3}-1|\lt\epsilon$, which proves that $\lim_{x\to2}h(x)=1=h(2)$
Step 6. We conclude that the function is continuous at $c=2$
