Answer
The proof is detailed in the Work step by step below.
Work Step by Step
$$g(x)=\frac{1}{2x}\hspace{1cm}c=\frac{1}{4}$$
For $g$ to be continuous at $c=1/4$, $\lim_{x\to 1/4}g(x)=g(1/4)=\frac{1}{2\times\frac{1}{4}}=\frac{1}{\frac{1}{2}}=2$
In other words, we need to prove that $\lim_{x\to 1/4}g(x)=2$.
According to the formal definition of limit, for any values of $\epsilon\gt0$, there must exist a corresponding value of $\delta\gt0$ such that for all $x$,
$$0\lt|x-1/4|\lt\delta\Rightarrow|g(x)-2|\lt\epsilon$$
Examining the inequality: $$|g(x)-2|\lt\epsilon$$ $$|\frac{1}{2x}-2|\lt\epsilon$$ $$-\epsilon\lt\frac{1}{2x}-2\lt\epsilon$$ $$2-\epsilon\lt \frac{1}{2x}\lt2+\epsilon$$ $$\frac{1}{2-\epsilon}\gt 2x\gt\frac{1}{2+\epsilon}$$ $$\frac{1}{4-2\epsilon}\gt x\gt\frac{1}{4+2\epsilon}$$
So if $x\in(\frac{1}{4+2\epsilon},\frac{1}{4-2\epsilon})$ then $|g(x)-2|\lt\epsilon$
Take $\delta$ to be the distance from $c=1/4$ to the nearer endpoint of $(\frac{1}{4+2\epsilon},\frac{1}{4-2\epsilon})$, or $\delta=\min(1/4-\frac{1}{4+2\epsilon},\frac{1}{4-2\epsilon}-1/4)$.
If $\delta$ has this value or any smaller values, then because $0\lt|x-1/4|\lt\delta$, $x$ will always lie in the interval $(\frac{1}{4+2\epsilon},\frac{1}{4-2\epsilon})$ so that $|g(x)-2|\lt\epsilon$.
Therefore, $\lim_{x\to1/4}g(x)=g(1/4)=2$. $g$ is continuous at $x=1/4$.
