$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational. Does a limit exist?
Does the following function have a limit as x approaches a?
$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.
Answers given in terms of delta-epsilon please!
My thoughts so far:
(1.) Its graph seems to show this function acting as though it were two constant functions
(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.
I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.
real-analysis
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
add a comment |
Does the following function have a limit as x approaches a?
$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.
Answers given in terms of delta-epsilon please!
My thoughts so far:
(1.) Its graph seems to show this function acting as though it were two constant functions
(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.
I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.
real-analysis
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50
add a comment |
Does the following function have a limit as x approaches a?
$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.
Answers given in terms of delta-epsilon please!
My thoughts so far:
(1.) Its graph seems to show this function acting as though it were two constant functions
(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.
I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.
real-analysis
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
Does the following function have a limit as x approaches a?
$f(x) = 0$ if x is irrational, $f(x) = 1$ if x is rational.
Answers given in terms of delta-epsilon please!
My thoughts so far:
(1.) Its graph seems to show this function acting as though it were two constant functions
(2.) Between any two rational numbers there are infinitely many rational numbers; between any two rational numbers there are infinitely many irrational numbers. So I cannot think of any interval where rational numbers wouldn't interrupt irrational numbers, or irrational numbers interrupt rational numbers.
I'm sorry to have so little to add to the question. This is just one of those functions which goes beyond anything you've learned.
real-analysis
real-analysis
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
edited Nov 24 at 19:43
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
asked Nov 24 at 18:28
Miguel Cumming-Romo
265
265
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50
add a comment |
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50
add a comment |
2 Answers
2
active
oldest
votes
You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.
Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.
answered Nov 24 at 18:36
user3482749
2,421414
add a comment |
Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.
Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.
With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).
answered Nov 24 at 18:56
Ovi
12.4k1038111
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011900%2ffx-0-if-x-is-irrational-fx-1-if-x-is-rational-does-a-limit-exist%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.
Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.
answered Nov 24 at 18:36
user3482749
2,421414
add a comment |
You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.
Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.
answered Nov 24 at 18:36
user3482749
2,421414
add a comment |
You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.
Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.
answered Nov 24 at 18:36
user3482749
2,421414
You are correct that there is no limit: in particular, you can find sequences $(x_n)$ and $(y_n)$ converging to $a$ such that all $x_n$ are rational and all $y_n$ are irrational (this follows from your density argument: simply pick some rational $x_0$ and irrational $y_0$, and construct your sequences by repeatedly choosing a rational/irrational from the interval centred on $a$ of length $|a - x_0|$ / $|a - y_0|$ (not including $a$): the sequences thus produced have the required properties.
Thus, for every potential limit $c$, choose $varepsilon = frac{1}{2}$. Then for any $delta > 0$, there is some $n$ such that $x_n$ and $y_n$ lie in $(a-delta,a+delta)setminus {a}$. Finally, notethat either $|f(x_n) - c| = |1 - c| geq varepsilon$ or $|f(y_n) - c| = |c| geq varepsilon$ (since if $|c| < frac{1}{2}$, then $c < frac{1}{2}$, so $|1 - c| = 1 - c geq frac{1}{2}$). Thus, $limlimits_{xto c}f(x)$ does not exist.
answered Nov 24 at 18:36
user3482749
2,421414
answered Nov 24 at 18:36
user3482749
2,421414
answered Nov 24 at 18:36
user3482749
2,421414
answered Nov 24 at 18:36
user3482749
2,421414
2,421414
add a comment |
add a comment |
Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.
Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.
With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).
answered Nov 24 at 18:56
Ovi
12.4k1038111
add a comment |
Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.
Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.
With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).
answered Nov 24 at 18:56
Ovi
12.4k1038111
add a comment |
Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.
Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.
With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).
answered Nov 24 at 18:56
Ovi
12.4k1038111
Intuitively, a function being continuous means that if you nudge the input just a little bit, the output also changes just a little bit.
Let's consider $a=5$. We know $f(5)=1$. Now if you nudge the input by a very small irrational number $c$, we get $f(5+c) = 0$, so the output, $0$, is relatively far from the output of our $a$ value. The nail in the coffin is that you can take $c$ to be as small as you want, and you will always get a relatively far output. This is exactly a property which discontinuous functions have.
With continuous functions on the other hand, if you make that $c$ smaller and smaller, the distance between the outputs should get smaller and smaller too, not stay the same (or get bigger).
answered Nov 24 at 18:56
Ovi
12.4k1038111
answered Nov 24 at 18:56
Ovi
12.4k1038111
answered Nov 24 at 18:56
Ovi
12.4k1038111
answered Nov 24 at 18:56
Ovi
12.4k1038111
12.4k1038111
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011900%2ffx-0-if-x-is-irrational-fx-1-if-x-is-rational-does-a-limit-exist%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Similar (math.stackexchange.com/questions/403338/…)
– Yadati Kiran
Nov 24 at 18:34
Consider for example $x=1$ and $x_n = 1 + frac{sqrt{2}}{n}$. Show that $x_n$ is irrational for any integer $n$. Apply the definition of continuity.
– Winther
Nov 24 at 19:50