A quick question. Is the function $\frac {x-2}{x-2}$ defined at $x=2$? For, when I plotted it, it was shown to be continuous. Also, if I try to prove continuity, the left limit $=1$ and right limit is also equal to $1$, but $f(2)$ takes the $0/0$ form. How do I prove the continuity?
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4To understand the continuity, you first have to define $f \left( 2 \right)$. What value do you want it to take? – Aniruddha Deshmukh Apr 10 '21 at 08:24
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See https://math.stackexchange.com/questions/1482787/can-we-talk-about-the-continuity-discontinuity-of-a-function-at-a-point-which-is/1482900#1482900 – Michael Hoppe Apr 10 '21 at 10:26
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Since $f(2)$ is not defined at $2$ since $0/0$ is undefined it makes no sense it ask if it’s continuous at $2$ . To make your function continuous at $2$ define $f(2)$ to be $1$ then $\lim_{x\to2} f(x)$ = $f(2)=1$ .
It might also help to understand the definition of a function if $X$ and $Y$ are sets and let $P(x,y)$ be a property for every object $x$ belonging to $X$ and $y$ belonging to $Y$ such that for every $x$ there is exactly one $y$ for which $P(x,y)$ is true . Then the function is defined by the property $P$ on the domain $X$ and range $Y$ is defined to be the object which given any input $x$ belonging to $X$ assigns an output $f(x)$ belonging to $Y$ .
From analysis 1 by Tao.
Vivaan Daga
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1You can, but in this case it is not continuous since $\lim_{x\to 2} f(x) \neq f(2) = 73$. – Rem Apr 10 '21 at 08:38
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A function consists of three pieces of data: a domain, a codomain, and an instruction on how to assign exactly one element of the codomain to every element of the domain. You only supplied us with one of those data pieces, the instruction. Domain and codomain are missing, so we are forced to guess them from context. The usual convention on how to guess is:
- to assume that you work in a subset of the real or complex numbers
- to assume that you follow the convention that complex inputs are written as $z$, while real inputs are written as $x$
- to assume that your codomain is the reals or the complex numbers, depending on the previous point
- to assume that you want the domain to be the largest subset of the number set you're working with where your given instruction makes sense
Since you wrote your instruction using $x$, I assume you're considering a function from a subset of the reals to the reals. Since your instruction makes sense for all real numbers except $2$, I assume that your domain is $\mathbb R\backslash\{2\}$. So $2$ is not in the domain of your function, so it can't be continuous or discontinuous at $2$. It just isn't anything at $2$, the same way it isn't anything at Paris: Paris is not part of the functions assumed domain, so it's neither continuous, nor discontinuous at Paris.
Notice that these are a lot of assumptions about your intentions. But then again, the question itself reveals that you yourself aren't quite sure about your intentions. If you intended for $2$ not to be in the domain, you probably wouldn't ask this question. Instead, you don't know wether you want $2$ to be in the domain. Now at this point, you have to pick, since you supplied the function. Is $2$ in the domain or not? If not, then go with what I wrote above. If yes, then you need to rephrase your question in one of two ways:
- You tell us what value you want the function to take at $x=2$, and keep the question "is it continuous at $2$?"
- You ask us wether there is any way at all to choose the value it takes at $2$ to make it continuous.
I can answer the second question: if the function maps $2$to $1$, then that would make it continuous, since then it's just the constant function $\mathbb R\to\mathbb R,~x\mapsto1$. Note how I specified domain, codomain, and instruction to avoid the kind of ambiguity I described above.
Also a note: if you see a function written the way you supplied it, you should make the same assumptions I made above. Most importantly, that the function's domain is only those numbers where the given instruction makes sense.
Vercassivelaunos
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No the function $f(x)=\frac {x-2 }{x-2}$ is not defined at $x=2$ and so not continuous at $x=2$. In fact, it’s not even discontinuous at $x=2$ as it is not defined at $x=2$. Note that domain of $f$ is $\mathbb R-\{2\}$.
However, you can define an extension $g:\mathbb R\to \mathbb R$ of $f$ to $\mathbb R$ as below:
$g(x)=\begin{cases} f(x)=1; x\ne 2\\\lim_{x\to 2} f(x)=1; x=2\end{cases}$
Notice that this extension of $f$ ,that is $g$ is now continuous everywhere.
Koro
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1"In fact, it’s not even discontinuous" I don't think I understand that. Would you care to explain? – Eisenstein Apr 10 '21 at 08:27
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@Eisenstein: Continuity or discontinuity is defined at a point where function is defined. The function you mentioned is not defined at x=2 simply because division by 0 is not defined. – Koro Apr 10 '21 at 08:29
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That actually depends on how discontinuity is defined. If defined in terms of punctured neighbourhoods, then technically this function would be discontinuous at $2$. So in a sense we’d be saying a function is discontinuous everywhere outside its domain – Chubby Chef Apr 10 '21 at 08:36
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2@Koro So to sum up, is the function neither continuous nor discontinuous? – Eisenstein Apr 10 '21 at 08:42
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1At $x=2$, yes! But now you see how we can (sometimes) extend $f$ to $g$ to make it continuous everywhere. – Koro Apr 10 '21 at 08:43
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