Function continuity $(x^2 - 1)/( x - 1)$ What can you say about the function continuity of $$\frac{x^2 - 1}{x - 1}$$ at $x = 1$?
There is an asymptote at $x = 1$, however the limit as $x$ goes to $1$ becomes $2$ because $f(x) = x + 1$ after factoring, also the limit becomes $2$ because we again factor, so in total by definition the function is continuous, however if we draw the graph then there is an asymptote at $x = 1$, so it's discontinuous.
So in then final result what can we say about the continuity of this function?
 A: That is some poor wording on this question.


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*$(x^2-1)/(x-1)$ is not a function as domain/codomain are missing. It should better be written as $$f:\mathbb R\setminus\{1\}\rightarrow\mathbb R,~x\mapsto f(x)=\frac{x^2-1}{x-1}.$$

*For a function to be continuous at some point $c$ of its domain, it is neccesary for the function to be defined at this point. As $f$ is not well-defined at $x=1$, it makes no sense to ask about continuity.


A possible rewording of the question: 

Let $f:\mathbb R\setminus\{1\}\rightarrow\mathbb R,~x\mapsto\frac{x^2-1}{x-1}$. Does a continuous extension exist e.g. a continuous function $$\tilde f:\mathbb R\rightarrow\mathbb R,~x\mapsto\begin{cases} f(x), & x\neq 1 \\ c, & x=1\end{cases}$$ with $c\in\mathbb R$?

Now: $f$ is continuous at a point $x_0$ iff for every sequence $(x_n)_{n\in\mathbb N}$ with $\lim\limits_{n\to\infty} x_n=x_0$ we also have $\lim\limits_{n\to\infty} f(x_n)=f(x_0)$. For functions $f:\mathbb R\rightarrow\mathbb R$ or $f:(a,b)\rightarrow \mathbb R$ with $-\infty<a<b<\infty$ this can be simplified to $$\lim\limits_{x\uparrow x_0}f(x)=f(x_0)=\lim\limits_{x\downarrow x_0} f(x).$$
Factorising is a good start, you have already done 90% of the work there is to do...can you take it from here?

On a sidenote: drawing the graph is something one can do to get a general idea of the function, but it will not tell you anything about the function being continuous or not. And I don't know how you drew the graph, but there won't be any asymptote at $x=1$.
