What Can We Say About the Continuity of $y=\frac{x}{x}$ at $x=0$?

Question

If we can't divide by $0$, should $\frac{x}{x}$ be discontinuous and undefined at $x=0$ or is it continuous with value $1$? Most online graph calculators plot a continuous curve.

If it's continuous at $x=0$ with $y=1$, then we should be able to say that $\frac{(a-b)}{(a-b)} = 1$ at $a=b$. Or $q^2*\frac{p}{q}$ is $0$ at $q = 0$. And that whole proof of $2=1$ would hold true. The graphs online for say $\frac{x^2-4}{x-2}$ at $x=2$ are puzzling me.

How would you expect an online graphing tool to show a discontinuity of this kind? — David K, Jul 13 '19 at 21:44
Show a circle or a dot with the center at the point like how we do on paper. Glad I asked this, had no idea about removable discontinuity, and that we are "patching" q^2*p/q as pq in our regular algebaic calculations, even though it's undefined at q=0. — whitewalker, Jul 13 '19 at 21:52
That’s what some textbooks do, but there’s a trade off: because of the blank area inside the circle, some of the points where the function is defined are not plotted. There could also be two discontinuities very close on the graph; do you overlap them, merge, or what? But I suspect the real reason is more related to the plotting algorithms: it’s a lot easier to find all the points you need in order to make a plot appear continuous where it should be, than to find all the removable discontinuities in an arbitrary function. — David K, Jul 14 '19 at 00:34
Who on Earth is voting to close this? This is a golden teaching opportunity and it was articulated well as a question. — The Count, Jul 14 '19 at 00:58

Saketh Malyala · Accepted Answer · 2019-07-21T08:59:29.573

Haha. That's what we call a removable discontinuity.

Now $f(x)=1$ and $g(x)=\displaystyle \frac{x}{x}$ are very similar functions, with one difference you just pointed out. (Discontinuity at $x=0$)

A similar thing for $f(x) = x-2$ and $g(x) = \displaystyle \frac{x^2-4}{x+2}$. (Discontinuity at $x=-2$.)

We notice that even at $x=-1.9999$ the values of the function are the same. We say that the limit of $g(x)$ as $x$ approaches $-2$ exists, even if, as we can see, $g(-2)$ does not exist.

A nonremovable discontinuity would be something like, $\displaystyle \frac{1}{x}$ as $x$ approaches $0$.

I hope that helps you understand. It is still a discontinuity, but since it's removable, we can sometimes patch it up.

Consider the piecewise function $h(x)$.

$h(x)=\displaystyle \begin{cases} \frac{x}{x} & x \neq 0 \\ 1 & x=0 \end{cases}$

We were only able to "patch up" the discontinuity because it was removable. It was a infinitesimally small hole, not an entire asymptote like in the case of $f(x) = \displaystyle \frac{1}{x}$.

@YiFan thanks for the catch. i think i was thinking about how the value of the function is $1$ when i wrote that — Saketh Malyala, Jul 14 '19 at 02:43

score 1 · Answer 2 · answered Jul 13 '19 at 21:23

1

$y=\frac{x}{x}$ is not defined at $x=0$, so it's not continuous there, but it has what is known as a "removable discontinuity" there. You can find many pages/videos explaining the idea. Here is what Wikipedia has to say.

answered Jul 13 '19 at 21:23

Mark S.

25,893

Close, but no cigar. As $y=x/x$ isn't defined at $x=0$ the sentence: "$y$ is continuous (or discontinuous) at $x=0$." is not a statement at all, i. e., it can be neither true nor false. – Michael Hoppe Jul 14 '19 at 10:36
@Michael This depends very sensitively on foundational choices. For instance, a number of calculus textbooks list three requirements for continuity at $a$ like: 1. Be defined at $a$. 2. Have a limit at $a$. 3. The limit equals the value of the function at $a$. For those books, we would absolutely say it's discontinuous because it fails to be defined. I agree this may conflict with some calculus/topology textbooks, but in a first calculus course I think it's reasonable since everything is basically a partial function $\mathbb R\nrightarrow \mathbb R$. – Mark S. Jul 14 '19 at 12:43
In case a function is not defined, it has no properties at all -- except being undefined, see https://math.stackexchange.com/questions/1482787/can-we-talk-about-the-continuity-discontinuity-of-a-function-at-a-point-which-is/1482900#1482900. Otherwise one could state the $f(x)=\sqrt x$, defined on the non-negative reals, is discontinuous at $x=-1$ ... – Michael Hoppe Jul 14 '19 at 12:59
@Michael In many calculus books we could indeed say just that about $\sqrt x$. Citing yourself isn't the ultimate authority since books/sources would disagree with each other about this. I can try to dig up some particular ones if you'd like. I don't know if your objection stems from a particular definition of function, but if so, I think we could allow for the books that disagree with you by using the concept of partial function from the reals instead. – Mark S. Jul 14 '19 at 13:13

Nitin Uniyal · Answer 3 · 2019-07-16T03:38:03.540

1

For continuity, you need $lim_{x\rightarrow 0^-} f(x)=lim_{x\rightarrow 0^+}f(x)=f(0)$.

$ lim_{x\rightarrow 0^-}\frac{x}{x}= lim_{x\rightarrow 0^+}\frac{x}{x}=1$, due to the fact that you can cancel $x$ in numerator and denominator as $x\ne 0$. But $f(0)$ is not defined specifically.

edited Jul 16 '19 at 03:38

answered Jul 14 '19 at 02:38

Nitin Uniyal

8,108

$f(0)$ is not defined specifically. But cancellation is valid as $x\ne 0$. – Nitin Uniyal Jul 14 '19 at 16:01
@N.Bar....Atleast you should have some valid reason to downvote. – Nitin Uniyal Jul 16 '19 at 03:41
why LHL doesn't exist? – Nitin Uniyal Jul 17 '19 at 00:03
Well. $x<0$ is negative real axis and$\lim_{x\rightarrow 0-} \frac{x}{x}=\lim_{h\rightarrow 0} \frac{0-h}{0-h}=1$. Hope it makes you clear. – Nitin Uniyal Jul 17 '19 at 01:19
The given function is a rational function $x/x$, not $\sqrt{x/x}$. I think u misunderstood the function. – Nitin Uniyal Jul 17 '19 at 01:24
Let us continue this discussion in chat. – N. Bar Jul 17 '19 at 01:25

score 0 · Answer 4 · answered Jul 14 '19 at 02:48

The rational function $y = \frac{x}{x}$ is continuous everywhere it's defined. As it is not defined at $x=0$ it cannot be continuous there. Hence, it is continuous everywhere except at $0$.

Some graphing calculators are notorious for giving incorrect graphs because they don't indicate where holes occur. In this case, the graph of $y = \frac{x}{x}$ will look like the graph of the constant function $y = 1$ everywhere, except at $0$, where there will be no point.

Matt Samuel · Answer 5 · 2019-07-13T21:43:36.037

-1

You can't divide by $0$, so the value at $0$ is undefined. The screens of graphing calculators have a finite resolution and although a hole should be in the graph, the hole might be "in between" two actual pixels and may not be visible. With high resolution graphs, even if it's displayed it may be too small to see. There are many reasons we don't rely on visuals to make deductions and this is one of them.

edited Jul 13 '19 at 21:43

answered Jul 13 '19 at 21:23

Matt Samuel

59,287

2

Wow, I had no idea that the pixels could be the reason why a hole isn't shown. Maybe calculators don't really take that stuff into account since they are numerical. Desmos shows holes though. – Jul 13 '19 at 21:27
The 'hole' only exists at $x=0$, so it's a single point and you shouldn't be able to see it at all. If your graph does show a hole, then that's down to the graphing calculator and nothing to do with the actual mathematical curve. – nickgard Jul 13 '19 at 21:58
@nick A pixel shows a single value of the function. If the $x$ coordinate is 0,that pixel won't be displayed. It just may be that there is no pixel with $x$ coordinate 0. – Matt Samuel Jul 13 '19 at 22:02
Not sure why this was downvoted, it's the only answer that addresses the graphs. – Matt Samuel Jul 14 '19 at 12:10

What Can We Say About the Continuity of $y=\frac{x}{x}$ at $x=0$?

5 Answers5