Show that $g(x)=[frac1x]sin x$ has a limit in $x=0$.












2












$begingroup$


Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



( $[1/x]$ as greatest integer less than or equal to $1/x$)



I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



I know :
$$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



And I know :



$$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



I just don’t know how to use this stuffs to solve the question.



Thanks in advance fo the help .






calculus limits floor-function







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



    ( $[1/x]$ as greatest integer less than or equal to $1/x$)



    I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
    I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



    I know :
    $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



    And I know :



    $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



    I just don’t know how to use this stuffs to solve the question.



    Thanks in advance fo the help .






    calculus limits floor-function







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



      ( $[1/x]$ as greatest integer less than or equal to $1/x$)



      I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
      I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



      I know :
      $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



      And I know :



      $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



      I just don’t know how to use this stuffs to solve the question.



      Thanks in advance fo the help .






      calculus limits floor-function







      share|cite|improve this question











      $endgroup$




      Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



      ( $[1/x]$ as greatest integer less than or equal to $1/x$)



      I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
      I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



      I know :
      $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



      And I know :



      $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



      I just don’t know how to use this stuffs to solve the question.



      Thanks in advance fo the help .






      calculus limits floor-function




      calculus limits floor-function






      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 3:37







      user98602

















      asked Dec 6 '18 at 18:18









      NegarNegar

      477




      477






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



          The function $sin$ is positive for small positive numbers and negative for small negative numbers.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



            Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              The OP is about the integer part [1/x].
              $endgroup$
              – gimusi
              Dec 6 '18 at 18:48










            • $begingroup$
              @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
              $endgroup$
              – J.G.
              Dec 6 '18 at 18:55



















            -3












            $begingroup$

            HINT



            You are mostly done, indeed since by floor function definition we have



            $$frac1x-1leleft[frac1xright]le frac1x$$



            then assuming wlog $0<x<1$



            $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



            and then use squeeze theorem.






            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%2f3028851%2fshow-that-gx-frac1x-sin-x-has-a-limit-in-x-0%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



              The function $sin$ is positive for small positive numbers and negative for small negative numbers.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                The function $sin$ is positive for small positive numbers and negative for small negative numbers.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                  The function $sin$ is positive for small positive numbers and negative for small negative numbers.






                  share|cite|improve this answer











                  $endgroup$



                  If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                  The function $sin$ is positive for small positive numbers and negative for small negative numbers.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 8:40









                  Martin Sleziak

                  44.8k10119273




                  44.8k10119273










                  answered Dec 6 '18 at 18:36









                  UserSUserS

                  1,5541112




                  1,5541112























                      1












                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55
















                      1












                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55














                      1












                      1








                      1





                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$



                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 6 '18 at 18:36









                      J.G.J.G.

                      29.1k22845




                      29.1k22845








                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55














                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55








                      1




                      1




                      $begingroup$
                      The OP is about the integer part [1/x].
                      $endgroup$
                      – gimusi
                      Dec 6 '18 at 18:48




                      $begingroup$
                      The OP is about the integer part [1/x].
                      $endgroup$
                      – gimusi
                      Dec 6 '18 at 18:48












                      $begingroup$
                      @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                      $endgroup$
                      – J.G.
                      Dec 6 '18 at 18:55




                      $begingroup$
                      @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                      $endgroup$
                      – J.G.
                      Dec 6 '18 at 18:55











                      -3












                      $begingroup$

                      HINT



                      You are mostly done, indeed since by floor function definition we have



                      $$frac1x-1leleft[frac1xright]le frac1x$$



                      then assuming wlog $0<x<1$



                      $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                      and then use squeeze theorem.






                      share|cite|improve this answer











                      $endgroup$


















                        -3












                        $begingroup$

                        HINT



                        You are mostly done, indeed since by floor function definition we have



                        $$frac1x-1leleft[frac1xright]le frac1x$$



                        then assuming wlog $0<x<1$



                        $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                        and then use squeeze theorem.






                        share|cite|improve this answer











                        $endgroup$
















                          -3












                          -3








                          -3





                          $begingroup$

                          HINT



                          You are mostly done, indeed since by floor function definition we have



                          $$frac1x-1leleft[frac1xright]le frac1x$$



                          then assuming wlog $0<x<1$



                          $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                          and then use squeeze theorem.






                          share|cite|improve this answer











                          $endgroup$



                          HINT



                          You are mostly done, indeed since by floor function definition we have



                          $$frac1x-1leleft[frac1xright]le frac1x$$



                          then assuming wlog $0<x<1$



                          $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                          and then use squeeze theorem.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 7 '18 at 8:49

























                          answered Dec 6 '18 at 18:39









                          gimusigimusi

                          92.9k84494




                          92.9k84494






























                              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%2f3028851%2fshow-that-gx-frac1x-sin-x-has-a-limit-in-x-0%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

                              mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                              Image
                              0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
                              Read more

                              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