Proving that an Epsilon-Delta Proof is not true












0












$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20
















0












$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20














0












0








0





$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$




We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta




proof-verification proof-explanation epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 12:25







Viserom

















asked Nov 30 '18 at 12:14









ViseromViserom

123




123












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20


















  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20
















$begingroup$
Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
$endgroup$
– Leo163
Nov 30 '18 at 12:20




$begingroup$
Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
$endgroup$
– Leo163
Nov 30 '18 at 12:20










2 Answers
2






active

oldest

votes


















2












$begingroup$

$$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



Follow the hint,



Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Given $delta>0$, we look for $x$ such that



    $$0<|x|<delta $$
    and
    $$|f(x)-frac 12|ge frac 14.$$



    $$iff$$



    $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



    so we can take $x_0=-frac{delta}{2}$
    with $f(x_0)=0$.






    share|cite|improve this answer









    $endgroup$













      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%2f3020025%2fproving-that-an-epsilon-delta-proof-is-not-true%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









      2












      $begingroup$

      $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



      Follow the hint,



      Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



        Follow the hint,



        Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



          Follow the hint,



          Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






          share|cite|improve this answer









          $endgroup$



          $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



          Follow the hint,



          Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 12:20









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118























              1












              $begingroup$

              Given $delta>0$, we look for $x$ such that



              $$0<|x|<delta $$
              and
              $$|f(x)-frac 12|ge frac 14.$$



              $$iff$$



              $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



              so we can take $x_0=-frac{delta}{2}$
              with $f(x_0)=0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Given $delta>0$, we look for $x$ such that



                $$0<|x|<delta $$
                and
                $$|f(x)-frac 12|ge frac 14.$$



                $$iff$$



                $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                so we can take $x_0=-frac{delta}{2}$
                with $f(x_0)=0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Given $delta>0$, we look for $x$ such that



                  $$0<|x|<delta $$
                  and
                  $$|f(x)-frac 12|ge frac 14.$$



                  $$iff$$



                  $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                  so we can take $x_0=-frac{delta}{2}$
                  with $f(x_0)=0$.






                  share|cite|improve this answer









                  $endgroup$



                  Given $delta>0$, we look for $x$ such that



                  $$0<|x|<delta $$
                  and
                  $$|f(x)-frac 12|ge frac 14.$$



                  $$iff$$



                  $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                  so we can take $x_0=-frac{delta}{2}$
                  with $f(x_0)=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 13:18









                  hamam_Abdallahhamam_Abdallah

                  38.1k21634




                  38.1k21634






























                      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%2f3020025%2fproving-that-an-epsilon-delta-proof-is-not-true%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

                      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

                      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