Is $\frac{x^2-1}{x-1}$ continuous at $x=1$? [duplicate]
Solution 1:
Does this mean this function is not continuous at $x=1$?
Yes, and you give the correct reason, that is, $f(1)$ does not exist.
However, $f$ can be extended in order to make it continuous:
$$f(x)=\begin{cases} \frac{x^2-1}{x-1} &\text{ if } x\neq 1 \\ 2 &\text{ if } x=1\end{cases}$$
is a continuous function and is equivalent to the function $f(x) = x+1\ (\forall x\in\mathbb{R})$.
Solution 2:
No, your function is not continuous at 1. It doesn't satisfy your definition for continuity at some a. Although the limit exists, $f(1)$ doesn't.