Answer
See explanations.
Work Step by Step
Step 1. Given the function $g(x)=\frac{x^2-2x-3}{2x-6}$, we need to prove the limit $\lim_{x\to3}f(x)$ exits.
Step 2. For any given small value $\epsilon\gt0$, we need to find a corresponding $\delta\gt0$ such that for all $x$ in the interval of $|x-3|\lt\delta$, we have $|g(x)-L|\lt\epsilon$, where $L$ is the limit.
Step 3. Since when $x\to3$, $x\ne3$, we have $\lim_{x\to3}g(x)=\lim_{x\to3}\frac{x^2-2x-3}{2x-6}=\lim_{x\to3}\frac{(x+1)(x-3)}{2(x-3)}=\lim_{x\to3}\frac{x+1}{2}$
Step 4. We need to prove the above limit $L=2$. With $|\frac{x+1}{2}-2|\lt\epsilon$, we have $-\epsilon\lt \frac{x+1}{2}-2\lt\epsilon$ or $4-2\epsilon\lt x+1\lt4+2\epsilon$ which in turn gives $-2\epsilon\lt x-3\lt2\epsilon$. Let $\delta=2\epsilon$, we have $-\delta\lt x-3\lt\delta$ and $|x-3|\lt\delta$.
Step 5. Working the above step backward, we can show that for any given small value $\epsilon\gt0$, we can find a corresponding $\delta=2\epsilon$ such that for all $x$ in the interval of $|x-3|\lt\delta$, we have $|g(x)-2|\lt\epsilon$,
Step 6. Let $g(3)=2$; the function can be extended to be continuous at $x=3$.
