Analyse Equicontinuity (uniform continuity) of a function $ frac{sin(x)}{x},; x > 0 $












0












$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55
















0












$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55














0












0








0





$begingroup$


How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity







share|cite|improve this question











$endgroup$




How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$



When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?






calculus continuity equicontinuity




calculus continuity equicontinuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 23:37









user376343

3,9584829




3,9584829










asked Dec 9 '18 at 17:31









R0xx0rZzzR0xx0rZzz

747




747








  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55














  • 1




    $begingroup$
    By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:39










  • $begingroup$
    I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
    $endgroup$
    – R0xx0rZzz
    Dec 9 '18 at 17:41












  • $begingroup$
    Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:55








1




1




$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39




$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39












$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41






$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41














$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55




$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55










2 Answers
2






active

oldest

votes


















1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07





















1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032652%2fanalyse-equicontinuity-uniform-continuity-of-a-function-frac-sinxx%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









1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07


















1












$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07
















1












1








1





$begingroup$

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.






share|cite|improve this answer









$endgroup$



This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.



Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 18:32









Christian BlatterChristian Blatter

175k8115327




175k8115327








  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07
















  • 1




    $begingroup$
    To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 19:07










1




1




$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07






$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07













1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06
















1












$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06














1












1








1





$begingroup$

If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.






share|cite|improve this answer











$endgroup$



If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 17:05

























answered Dec 10 '18 at 0:10









zhw.zhw.

74.5k43175




74.5k43175












  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06


















  • $begingroup$
    typo corrected$,$
    $endgroup$
    – zhw.
    Dec 24 '18 at 17:06
















$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06




$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032652%2fanalyse-equicontinuity-uniform-continuity-of-a-function-frac-sinxx%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more