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