Is $f(x)=\frac{x^2-1}{x+1}$ continuous at $x^*=-1$? 
Is $f(x)=\frac{x^2-1}{x+1}$ continuous at $x^*=-1$?

For $f(x)=\frac{x^2-1}{x+1}$ there exists $\lim\limits_{x\to -1^-}f(x)=-2$ and  $\lim\limits_{x\to -1^+}f(x)=-2$ but it is not continuous at $x=-1$, but that would be a contradiction to the definition of continuity which states that for every function where both $\lim_{x\to x^{*-}}=\lim_{x\to x^{*+}}$ are equal then the function is continuous at the point $x^*$. What is wrong with my thoughts? (Or do I remember that incorrectly?)
 A: For a function to be continuous at a point, it must first be defined there. Since the expression for your function is not defined when $x=-1$, it cannot be continuous there.
However, the function 
$$
g(x)=\begin{cases}
f(x),&x\ne -1\\
-2,& x=-1
\end{cases}
$$
is continuous (we used existence of the limit to fill in the "hole" in the domain forming a continuous function on $\mathbb{R}$).
A: It is not true that $f$ is discontinuous at $-1$. Since $-1$ does not belong to the domain of $f$ (which I assume to be $\mathbb R\setminus\{-1\}$), the function $f$ is neither continuous nor discontinuous at that point. The concept of continuity is defined for points of the domain of the function and only for those points.
However, asserting that the limits $\lim_{x\to1^-}f(x)$ and $\lim_{x\to1^+}f(x)$ exist and are equal is equivalent to the assertion that you can extend $f$ to a continuous function from $\mathbb R$ into $\mathbb R$.
A: Your definition of continuity is not quite right. It's not simply that the left- and right-hand limits equal each other.
Instead, it is that the full limit exists (requiring the left- and right-hand limits to equal each other) and that limit has to equal the function's value at that point. In this case, there is no function value at that point.
