Answer
See explanations.
Work Step by Step
Step 1. To prove that the function is continuous at $c=1/4$, we need to show that the limit of the function $g(x)=1/(2x)$ exists at this point and is equal to a value of $g(1/4)=1/(2/4)=2$.
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 $|x-1/4|\lt\delta$, we have $|\frac{1}{2x}-2|\lt\epsilon$,
Step 3. The last inequality can be written as $-\epsilon\lt \frac{1}{2x}-2\lt\epsilon$ or $2-\epsilon\lt \frac{1}{2x}\lt\epsilon+2$ and $\frac{1}{4+2\epsilon} \lt x\lt \frac{1}{4-2\epsilon}$ (assuming $\epsilon\ne2$); subtract $\frac{1}{4}$ to get $\frac{1}{4+2\epsilon}-\frac{1}{4} \lt x-\frac{1}{4}\lt \frac{1}{4-2\epsilon}-\frac{1}{4}$
Step 4 Choose $\delta$ to be the smaller of $|\frac{1}{4+2\epsilon}-\frac{1}{4}|$ and $|\frac{1}{4-2\epsilon}-\frac{1}{4}|$ (if $\epsilon=2$, choose $\delta=|\frac{1}{4+2\epsilon}-\frac{1}{4}|=\frac{1}{8}$)
Step 5. With the chosen $\delta$ value, working the above steps backwards, we have for all $x$ in the interval $|x-1/4|\lt\delta$; we have $|\frac{1}{2x}-2|\lt\epsilon$, which proves that $\lim_{x\to1/4}g(x)=2=g(1/4)$
Step 6. We conclude that the function is continuous at $c=1/4$
