Calculate $lim_{xto 0}left(frac{3sin(x)-3x cos(x)}{x^3}right)^frac{1}{x}$












2












$begingroup$


This is an exercise from today's exam and I think I did something wrong:



Calculate the limit:



$$lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}.$$



To tackle this problem I used equivalent infinitesimals: $1-cos(x) sim frac{x^2}{2}$ as $(xto 0)$ and $sin(x)sim x$ as $(xto 0)$.



Then our limit:



$$begin{align}
lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}&=lim_{xto 0}left(frac{3x-3xcos(x)}{x^3}right)^{1/x}\
&=lim_{xto 0}left(frac{3x(1-cos(x))}{x^3}right)^{1/x}\
&=lim_{xto 0}left(frac{3x frac{x^2}{2}}{x^3}right)^{1/x} \
&=lim_{xto 0}left(frac{3}{2}right)^{1/x}
end{align}$$



Now I split the limit to see what the effect of negative $x$ will be:



$lim_{xto 0^+}left(frac{3}{2}right)^{1/x}=+infty$



$lim_{xto 0^-}left(frac{3}{2}right)^{1/x}=0$



I graphed the function and the limit seems to be 1 from both sides.






calculus limits taylor-expansion infinitesimals







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    This is an exercise from today's exam and I think I did something wrong:



    Calculate the limit:



    $$lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}.$$



    To tackle this problem I used equivalent infinitesimals: $1-cos(x) sim frac{x^2}{2}$ as $(xto 0)$ and $sin(x)sim x$ as $(xto 0)$.



    Then our limit:



    $$begin{align}
    lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}&=lim_{xto 0}left(frac{3x-3xcos(x)}{x^3}right)^{1/x}\
    &=lim_{xto 0}left(frac{3x(1-cos(x))}{x^3}right)^{1/x}\
    &=lim_{xto 0}left(frac{3x frac{x^2}{2}}{x^3}right)^{1/x} \
    &=lim_{xto 0}left(frac{3}{2}right)^{1/x}
    end{align}$$



    Now I split the limit to see what the effect of negative $x$ will be:



    $lim_{xto 0^+}left(frac{3}{2}right)^{1/x}=+infty$



    $lim_{xto 0^-}left(frac{3}{2}right)^{1/x}=0$



    I graphed the function and the limit seems to be 1 from both sides.






    calculus limits taylor-expansion infinitesimals







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      This is an exercise from today's exam and I think I did something wrong:



      Calculate the limit:



      $$lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}.$$



      To tackle this problem I used equivalent infinitesimals: $1-cos(x) sim frac{x^2}{2}$ as $(xto 0)$ and $sin(x)sim x$ as $(xto 0)$.



      Then our limit:



      $$begin{align}
      lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}&=lim_{xto 0}left(frac{3x-3xcos(x)}{x^3}right)^{1/x}\
      &=lim_{xto 0}left(frac{3x(1-cos(x))}{x^3}right)^{1/x}\
      &=lim_{xto 0}left(frac{3x frac{x^2}{2}}{x^3}right)^{1/x} \
      &=lim_{xto 0}left(frac{3}{2}right)^{1/x}
      end{align}$$



      Now I split the limit to see what the effect of negative $x$ will be:



      $lim_{xto 0^+}left(frac{3}{2}right)^{1/x}=+infty$



      $lim_{xto 0^-}left(frac{3}{2}right)^{1/x}=0$



      I graphed the function and the limit seems to be 1 from both sides.






      calculus limits taylor-expansion infinitesimals







      share|cite|improve this question











      $endgroup$




      This is an exercise from today's exam and I think I did something wrong:



      Calculate the limit:



      $$lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}.$$



      To tackle this problem I used equivalent infinitesimals: $1-cos(x) sim frac{x^2}{2}$ as $(xto 0)$ and $sin(x)sim x$ as $(xto 0)$.



      Then our limit:



      $$begin{align}
      lim_{xto 0}left(frac{3sin(x)-3xcos(x)}{x^3}right)^{1/x}&=lim_{xto 0}left(frac{3x-3xcos(x)}{x^3}right)^{1/x}\
      &=lim_{xto 0}left(frac{3x(1-cos(x))}{x^3}right)^{1/x}\
      &=lim_{xto 0}left(frac{3x frac{x^2}{2}}{x^3}right)^{1/x} \
      &=lim_{xto 0}left(frac{3}{2}right)^{1/x}
      end{align}$$



      Now I split the limit to see what the effect of negative $x$ will be:



      $lim_{xto 0^+}left(frac{3}{2}right)^{1/x}=+infty$



      $lim_{xto 0^-}left(frac{3}{2}right)^{1/x}=0$



      I graphed the function and the limit seems to be 1 from both sides.






      calculus limits taylor-expansion infinitesimals




      calculus limits taylor-expansion infinitesimals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 17 at 23:12









      Larry

      2,41331129




      2,41331129










      asked Jan 17 at 21:59









      user605734 MBSuser605734 MBS

      3029




      3029






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          $$3sin{x}-3xcos{x}=3left(x-frac{x^3}{6}+frac{x^5}{120}-...right)-3xleft(1-frac{x^2}{2}+frac{x^4}{24}-...right)=$$
          $$=x^3-frac{1}{10}x^5+...,$$
          which says that your limit is equal to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didnt really want the answer, I wanted to know where is my procedure flawed
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:14










          • $begingroup$
            @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:16












          • $begingroup$
            Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:28










          • $begingroup$
            Yes, of course, it's works. But you need to take the right approximation.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:30










          • $begingroup$
            Well I understand your answer but still I haven't found where my mistake lies
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:42











          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%2f3077567%2fcalculate-lim-x-to-0-left-frac3-sinx-3x-cosxx3-right-frac1x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $$3sin{x}-3xcos{x}=3left(x-frac{x^3}{6}+frac{x^5}{120}-...right)-3xleft(1-frac{x^2}{2}+frac{x^4}{24}-...right)=$$
          $$=x^3-frac{1}{10}x^5+...,$$
          which says that your limit is equal to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didnt really want the answer, I wanted to know where is my procedure flawed
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:14










          • $begingroup$
            @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:16












          • $begingroup$
            Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:28










          • $begingroup$
            Yes, of course, it's works. But you need to take the right approximation.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:30










          • $begingroup$
            Well I understand your answer but still I haven't found where my mistake lies
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:42
















          1












          $begingroup$

          $$3sin{x}-3xcos{x}=3left(x-frac{x^3}{6}+frac{x^5}{120}-...right)-3xleft(1-frac{x^2}{2}+frac{x^4}{24}-...right)=$$
          $$=x^3-frac{1}{10}x^5+...,$$
          which says that your limit is equal to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didnt really want the answer, I wanted to know where is my procedure flawed
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:14










          • $begingroup$
            @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:16












          • $begingroup$
            Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:28










          • $begingroup$
            Yes, of course, it's works. But you need to take the right approximation.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:30










          • $begingroup$
            Well I understand your answer but still I haven't found where my mistake lies
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:42














          1












          1








          1





          $begingroup$

          $$3sin{x}-3xcos{x}=3left(x-frac{x^3}{6}+frac{x^5}{120}-...right)-3xleft(1-frac{x^2}{2}+frac{x^4}{24}-...right)=$$
          $$=x^3-frac{1}{10}x^5+...,$$
          which says that your limit is equal to $1$.






          share|cite|improve this answer









          $endgroup$



          $$3sin{x}-3xcos{x}=3left(x-frac{x^3}{6}+frac{x^5}{120}-...right)-3xleft(1-frac{x^2}{2}+frac{x^4}{24}-...right)=$$
          $$=x^3-frac{1}{10}x^5+...,$$
          which says that your limit is equal to $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 22:08









          Michael RozenbergMichael Rozenberg

          105k1892198




          105k1892198












          • $begingroup$
            I didnt really want the answer, I wanted to know where is my procedure flawed
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:14










          • $begingroup$
            @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:16












          • $begingroup$
            Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:28










          • $begingroup$
            Yes, of course, it's works. But you need to take the right approximation.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:30










          • $begingroup$
            Well I understand your answer but still I haven't found where my mistake lies
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:42


















          • $begingroup$
            I didnt really want the answer, I wanted to know where is my procedure flawed
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:14










          • $begingroup$
            @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:16












          • $begingroup$
            Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:28










          • $begingroup$
            Yes, of course, it's works. But you need to take the right approximation.
            $endgroup$
            – Michael Rozenberg
            Jan 17 at 22:30










          • $begingroup$
            Well I understand your answer but still I haven't found where my mistake lies
            $endgroup$
            – user605734 MBS
            Jan 17 at 22:42
















          $begingroup$
          I didnt really want the answer, I wanted to know where is my procedure flawed
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:14




          $begingroup$
          I didnt really want the answer, I wanted to know where is my procedure flawed
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:14












          $begingroup$
          @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
          $endgroup$
          – Michael Rozenberg
          Jan 17 at 22:16






          $begingroup$
          @user605734 MBS You made mistake, when calculated the coefficient before $x^3$. You need to take more addend in the Taylor series for $sin$.
          $endgroup$
          – Michael Rozenberg
          Jan 17 at 22:16














          $begingroup$
          Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:28




          $begingroup$
          Is the method of changing sin(x) and 1-cos(x) with their respective equivalent infinitesimals valid in this limit?
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:28












          $begingroup$
          Yes, of course, it's works. But you need to take the right approximation.
          $endgroup$
          – Michael Rozenberg
          Jan 17 at 22:30




          $begingroup$
          Yes, of course, it's works. But you need to take the right approximation.
          $endgroup$
          – Michael Rozenberg
          Jan 17 at 22:30












          $begingroup$
          Well I understand your answer but still I haven't found where my mistake lies
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:42




          $begingroup$
          Well I understand your answer but still I haven't found where my mistake lies
          $endgroup$
          – user605734 MBS
          Jan 17 at 22:42


















          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%2f3077567%2fcalculate-lim-x-to-0-left-frac3-sinx-3x-cosxx3-right-frac1x%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

          MongoDB - Not Authorized To Execute Command

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
          Read more

          How to fix TextFormField cause rebuild widget in Flutter

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
          Read more

          in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

          Image
          1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
          Read more