about limit of ridichlet function

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hyourinn

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f(x) =1 if x is rational​
=0 if x is irrational​

no matter what delta i choose i still can find find both rational and irrational number that satisfy d<0 and 0<x<d
the problem is how i should choose epsilon regarding this function, frankly this obvious kind of function with only two possibilities of value is really hard for me to understand

basically i what i understand is to turn the problem into delta-epsilon language and see if there any contradiction within the process

|f(x)-L|<E for E>0 and d>0 that satisfy 0<x<d
​
|1-L|<E for rational x
|0-L|<E for irrational x

several times i saw the example about proof of dirichlet function, they're always use E=½ that seriously bugging me. so i tried various value for E and of course i will bring consequences for both delta and x, but since i can find infinite number of both irrational and rational number even within very small range i don't think much about it, since it quite different than "normal" function that really have strict epsilon-delta relation

so i've come with ,whenever i choose E with range 0<E<½ i will find such contradiction that i'm looking for(as a proof no limit). let's say E is equal to 0.2
the form above will turn into


|1-L|<0.2 for rational x
|0-L|<0.2 for irrational x

the bottom part (|0-L|<0.2) will result a contradiction, and that is what i'm looking for .

but when i try E within range ½<E<1
i cannot show any contradiction within it
let's say E 0.51




|1-L|<0.51 for rational x
|0-L|<0.51 for irrational x

whatever L i choose there will be no contradiction in it

so my point is

first,how exactly to choose epsilon in unique function like this(since this function only have 2 possible value and the maximum value is mere 1) ?

second it's common that dirichlet function is discontinuous everywhere(has no 2side limit) so why when i choose epsilon that higher than ½ i can't show the discontinue of the function?
isn't it supposed discontinuous everywhere?



i'm aware that i seems have some grave misunderstandings regarding the very basic definition of limit so i would really happy if someone can help me get past this

thank you in advance
 
Definition of a limit (L) at point X: for _any_ epsilon there _exists_ a small enough neighborhood of X where _all_ f(x) values are within epsilon from L.
How do we show that a limit does not exist at X? We find _one_ counter example.
I.e. there _exist_ an epsilon, such that for _any_ neighborhood around X there _exists_ an x where f(x) is farther away from L than epsilon.

There is nothing special about 1/2, other than it's half way between 0 and 1. Pick something else. E.g. 1/3.1415926. Or 1/2019.
No matter how close to X we get there will be x values where f(x) is more than 1/2019 away from 0 or from 1. So 0 and 1 can't be limits.
 
hyourinn said:
f(x) =1 if x is rational​
​
=0 if x is irrational​
no matter what delta i choose i still can find find both rational and irrational number that satisfy d<0 and 0<x<d
the problem is how i should choose epsilon regarding this function, frankly this obvious kind of function with only two possibilities of value is really hard for me to understand
Click to expand...
What limit are you trying to solve? What is the exact question?
 
lev888 said:
Definition of a limit (L) at point X: for _any_ epsilon there _exists_ a small enough neighborhood of X where _all_ f(x) values are within epsilon from L.
How do we show that a limit does not exist at X? We find _one_ counter example.
I.e. there _exist_ an epsilon, such that for _any_ neighborhood around X there _exists_ an x where f(x) is farther away from L than epsilon.

There is nothing special about 1/2, other than it's half way between 0 and 1. Pick something else. E.g. 1/3.1415926. Or 1/2019.
No matter how close to X we get there will be x values where f(x) is more than 1/2019 away from 0 or from 1. So 0 and 1 can't be limits.
Click to expand...
thank you very much , so the whole point is: say x= pi-3 and we prove that lim x → pi-3 is not equal with f(pi-3)=0 ?
in short if the f(x) of irrational number is 0 we prove that it limit is not 0? since the if both y and the limit display different value it mean there is jump in graph, it is something like that ?
 
Last edited:
Ah. The Dirichlet Function. \(\displaystyle f(x)=\begin{cases}1 &: x\in\mathbb{Q} \\ 0 &: x\notin\mathbb{Q}\end{cases}\)
Between any two real numbers there is a rational number and there is an irrational number. For sake on convenience let \(\displaystyle \mathscr{I}\) be the set of irrationals.
If \(\displaystyle \varepsilon>0 \) then \(\displaystyle \left[\exists \alpha\in\mathbb{Q}\cap\left(\pi^{-3}-\varepsilon,\pi^{-3}+\varepsilon\right)\right] \) AND \(\displaystyle \left[\exists \beta\in\mathscr{I}\cap\left(\pi^{-3}-\varepsilon,\pi^{-3}+\varepsilon\right)\right] \) thus \(\displaystyle |f(\alpha)-f(\beta)|=1\) so that means the image of any \(\displaystyle \varepsilon \text{-neighborhood}\) of \(\displaystyle \frac{1}{\pi^3}\) contains values of \(\displaystyle f\) that are one unit apart. That violates the idea of a limit
If you follow the link I provided to the Dirichlet Function you will find other similar functions that are nowhere continuous.

BTW: It should be with a capital D being a last name.
 
i'm sorry but i adding one more question
when i first time learn about definition of limit days ago the formal statement should be
|f(x)-L|<E but i just watch a video and it turn out
|f(x)-f(a)|<E and if f(x) is irrational so f(a) is rational and vice versa , and by doing this surely with absolute value the leftside of inequality will turn into 1 and it proof the contradiction regardless of E as long 0<E<1 or even 1

so from where |f(x)-f(a)|<E is obtained ?
 
hyourinn said:
i'm sorry but i adding one more question when i first time learn about definition of limit days ago the formal statement should be |f(x)-L|<E but i just watch a video and it turn out |f(x)-f(a)|<E and if f(x) is irrational so f(a) is rational and vice versa , and by doing this surely with absolute value the leftside of inequality will turn into 1 and it proof the contradiction regardless of E as long 0<E<1 or even 1 so from where |f(x)-f(a)|<E is obtained ?
Click to expand...
If you want to learn limits here is a free calculus textbook: In Elementary Calculus: An Infinitesimal Approach
The chapters and whole book is a free down-load. Here is the basic idea:
\(\displaystyle \mathop {\lim }\limits_{x \to c} f(x) = L\) means that if \(\displaystyle x\approx c\) (\(\displaystyle x\text{ is close to }c)\) then \(\displaystyle f(x)\approx L\) (\(\displaystyle f(x)\text{ is close to} L)\).
Now of course "close to" is not a mathematical concept. So we try to make it into one.
If \(\displaystyle \varepsilon>0\) then \(\displaystyle (L-\varepsilon,L+\varepsilon)\) is an open interval with \(\displaystyle L\) as its centre and \(\displaystyle \bf\varepsilon\) as its radius.
Clearly if \(\displaystyle \varepsilon\) is made smaller the set \(\displaystyle (L-\varepsilon,L+\varepsilon)\) is also "smaller" .
We look for \(\displaystyle \delta>0\) (which depends upon \(\displaystyle \varepsilon\)), so that if \(\displaystyle x\in(c-\delta,c+\delta)\) then \(\displaystyle f(x)\in(L-\varepsilon,L+\varepsilon)\).
Lets encode those sets. \(\displaystyle f(x)\in(L-\varepsilon,L+\varepsilon):~|f(x)-L|<\varepsilon\) and \(\displaystyle x\in(c-\delta,c+\delta):~|x-c|<\delta\)
Now I will do one and only one example: Prove \(\displaystyle \mathop {\lim }\limits_{x \to 2} ({x^3} - x) = 6\)
Given \(\displaystyle \varepsilon>0\), we have absolutely no control over \(\displaystyle \varepsilon\) it is fixed so we build a \(\displaystyle \delta>0\).
We start where we want to go: \(\displaystyle |x^3-x-6|\\|x^3-8-x+2|\\|(x-2)(x^2+2x+4)-(x-2)|\\|x-2|~|x^2+2x+3|\)
You see I built the factor \(\displaystyle |x-2|\) over which I have control in the \(\displaystyle \delta\).
Lets start with \(\displaystyle \delta=1\) so that
\(\displaystyle |x-2|<1\\-1<x-2<1\\1<x<4\)
If \(\displaystyle 1<x<4\) then \(\displaystyle 1<x^2<14\\2<2x<8\\7<x^2+2x+4<28\\|x^2+2x+4|<28\)
Now we have done it! let \(\displaystyle \delta=\min\left\{1,\frac{\varepsilon}{28}\right\}\)
\(\displaystyle |x - 2| < \delta \Rightarrow |f(x) - 6| < \varepsilon \).

Please do not ask for an explanation of any of that. The only way to learn to do those proofs is to do many of them.
Here are some to start on.
\(\displaystyle \begin{array}{l}a)\;\mathop {\lim }\limits_{x \to 5} \;2x = 10\\b)\;\mathop {\lim }\limits_{x \to 5} \;{x^2} = 25\\c)\;\mathop {\lim }\limits_{x \to 5} \;{x^2} - x = 20\\d)\;\mathop {\lim }\limits_{x \to 5} \;{x^3} = 125\\e)\;\mathop {\lim }\limits_{x \to 5} \;{x^3} - {x^2} = 100\end{array}\)

 
hyourinn said:
i'm sorry but i adding one more question
when i first time learn about definition of limit days ago the formal statement should be
|f(x)-L|<E but i just watch a video and it turn out
|f(x)-f(a)|<E and if f(x) is irrational so f(a) is rational and vice versa , and by doing this surely with absolute value the leftside of inequality will turn into 1 and it proof the contradiction regardless of E as long 0<E<1 or even 1

so from where |f(x)-f(a)|<E is obtained ?
Click to expand...
Probably |f(x)-L|<E and |f(x)-f(a)|<E just relate to different questions. L would be the limit of f as x approaches a; you can only replace that with f(a) if f is defined at a, and is continuous at a. So that video must have been about continuity, not limits in general.

In particular, it doesn't apply to your question about whether the Dirichlet function has a limit at any value x=a. It would apply if your question was about whether the function is continuous everywhere. You never actually stated the actual wording of the problem, but only, "it's about showing there is no limit in dirichlet function", so I assume it is the former, and there is no need to work with f(a).

But if the question were about continuity, you couldn't say that for any a and x, one being irrational implies the other is rational. (For this function both f(x) and f(a) are rational, always being either 0 or 1, so I'm assuming you meant x and a, not f(x) and f(a).)
 
Dr.Peterson said:
Probably |f(x)-L|<E and |f(x)-f(a)|<E just relate to different questions. L would be the limit of f as x approaches a; you can only replace that with f(a) if f is defined at a, and is continuous at a. So that video must have been about continuity, not limits in general.

In particular, it doesn't apply to your question about whether the Dirichlet function has a limit at any value x=a. It would apply if your question was about whether the function is continuous everywhere. You never actually stated the actual wording of the problem, but only, "it's about showing there is no limit in dirichlet function", so I assume it is the former, and there is no need to work with f(a).

But if the question were about continuity, you couldn't say that for any a and x, one being irrational implies the other is rational. (For this function both f(x) and f(a) are rational, always being either 0 or 1, so I'm assuming you meant x and a, not f(x) and f(a).)
Click to expand...
yes, thank you at least after watching it i feel i get better understanding now , but.. well please excuse me if i'm wrong but , if a function is not continous everywhere( "jumping" everywhere) isn't it proper to say that "there is no limit in this function".of course what i referring limit is about 2side limit, so what's wrong if we show that there is no limit in a function with showing it's discontinuous everywhere?
 
Your statement "there is no limit in this function" is just a little too vague. What you probably mean is that the function has no limit anywhere; and since that is true, it is not continuous anywhere. Certainly these are very closely connected; but saying that a function is not continuous is not identical to saying that it has no limit. When you want to prove something, you have to precisely state what you want to prove.
 
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