Function continuity $(x^2 - 1)/( x - 1)$

Question

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?

Note that if $f$ is not defined at that point, one cannot talk about the continuity at that point. — , Oct 12 '15 at 12:55
"if we draw the graph then there is an asymptote at x=1": this isn't true. — , Oct 12 '15 at 13:16
The graph of this function is the line $y = x + 1$ with a hole at the point $(1, 2)$. — N. F. Taussig, Oct 12 '15 at 15:12
Why is there a hole at (1, 2)? After all, the function sinc(x) = sin(x)/x is smooth and well defined at x=0... — Powers of Two, Feb 02 '16 at 12:09
Possible duplicate of Why does factoring eliminate a hole in the limit? — Alex M., Dec 03 '16 at 14:15

score 1 · Answer 1 · answered Oct 12 '15 at 13:10

That is some poor wording on this question.

$(x^2-1)/(x-1)$ 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 $x=1$.

Function continuity $(x^2 - 1)/( x - 1)$

1 Answers1

Linked

Related