Limit $limlimits_{xrightarrow 5}x^2=25$












2














From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







share|cite|improve this question




















  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27
















2














From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







share|cite|improve this question




















  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27














2












2








2







From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







share|cite|improve this question















From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta




calculus limits epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 12:14









Martin Sleziak

44.7k7115270




44.7k7115270










asked Sep 8 '14 at 19:21









Gᴇᴏᴍᴇᴛᴇʀ

4752621




4752621








  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27














  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27








1




1




Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
– almagest
Sep 8 '14 at 19:24




Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
– almagest
Sep 8 '14 at 19:24












Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
– TonyK
Sep 8 '14 at 19:34




Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
– TonyK
Sep 8 '14 at 19:34












I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
– Alex
Sep 8 '14 at 21:27




I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
– Alex
Sep 8 '14 at 21:27










2 Answers
2






active

oldest

votes


















0














Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
$$
|x^2 - 5| < 11|x - 5| < 10^{-3}
$$
For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






share|cite|improve this answer





























    0














    If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



    For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






    share|cite|improve this answer





















      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%2f924071%2flimit-lim-limits-x-rightarrow-5x2-25%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









      0














      Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
      $$
      |x^2 - 5| < 11|x - 5| < 10^{-3}
      $$
      For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






      share|cite|improve this answer


























        0














        Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
        $$
        |x^2 - 5| < 11|x - 5| < 10^{-3}
        $$
        For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






        share|cite|improve this answer
























          0












          0








          0






          Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
          $$
          |x^2 - 5| < 11|x - 5| < 10^{-3}
          $$
          For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






          share|cite|improve this answer












          Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
          $$
          |x^2 - 5| < 11|x - 5| < 10^{-3}
          $$
          For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 8 '14 at 19:26









          Omnomnomnom

          126k788176




          126k788176























              0














              If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



              For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






              share|cite|improve this answer


























                0














                If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






                share|cite|improve this answer
























                  0












                  0








                  0






                  If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                  For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






                  share|cite|improve this answer












                  If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                  For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 12:27









                  Mostafa Ayaz

                  13.7k3836




                  13.7k3836






























                      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.





                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f924071%2flimit-lim-limits-x-rightarrow-5x2-25%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

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more

                      web3.py web3.isConnected() returns false always

                      Image
                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                      Read more