Compute the limit or show it doesn't exist: $lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n})$












0












$begingroup$



Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$




I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.






calculus limits trigonometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    use the mean value theorem
    $endgroup$
    – user124910
    Dec 18 '18 at 4:05






  • 1




    $begingroup$
    $sinalpha-sinbeta=ldots$
    $endgroup$
    – Artem
    Dec 18 '18 at 4:10
















0












$begingroup$



Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$




I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.






calculus limits trigonometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    use the mean value theorem
    $endgroup$
    – user124910
    Dec 18 '18 at 4:05






  • 1




    $begingroup$
    $sinalpha-sinbeta=ldots$
    $endgroup$
    – Artem
    Dec 18 '18 at 4:10














0












0








0


1



$begingroup$



Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$




I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.






calculus limits trigonometry







share|cite|improve this question











$endgroup$





Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$




I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.






calculus limits trigonometry




calculus limits trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 4:24









tatan

5,82462760




5,82462760










asked Dec 18 '18 at 4:02









J. LastinJ. Lastin

1367




1367








  • 2




    $begingroup$
    use the mean value theorem
    $endgroup$
    – user124910
    Dec 18 '18 at 4:05






  • 1




    $begingroup$
    $sinalpha-sinbeta=ldots$
    $endgroup$
    – Artem
    Dec 18 '18 at 4:10














  • 2




    $begingroup$
    use the mean value theorem
    $endgroup$
    – user124910
    Dec 18 '18 at 4:05






  • 1




    $begingroup$
    $sinalpha-sinbeta=ldots$
    $endgroup$
    – Artem
    Dec 18 '18 at 4:10








2




2




$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05




$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05




1




1




$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10




$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10










4 Answers
4






active

oldest

votes


















2












$begingroup$

Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$




Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$







share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $sin x < x $
    $|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$



      For real $y,-1lecos yle1$



      For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        By the mean value theorem,



        $|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.






        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%2f3044759%2fcompute-the-limit-or-show-it-doesnt-exist-lim-n-to-infty-sin-sqrtn1%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
          $$|sin x - sin y| leq |x - y| .$$




          Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$







          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
            $$|sin x - sin y| leq |x - y| .$$




            Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$







            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
              $$|sin x - sin y| leq |x - y| .$$




              Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$







              share|cite|improve this answer









              $endgroup$



              Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
              $$|sin x - sin y| leq |x - y| .$$




              Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$








              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 18 '18 at 4:18









              TravisTravis

              63.7k769151




              63.7k769151























                  2












                  $begingroup$

                  $sin x < x $
                  $|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    $sin x < x $
                    $|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $sin x < x $
                      $|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$






                      share|cite|improve this answer









                      $endgroup$



                      $sin x < x $
                      $|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 18 '18 at 4:16









                      kayushkayush

                      2,215826




                      2,215826























                          0












                          $begingroup$

                          $$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$



                          For real $y,-1lecos yle1$



                          For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            $$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$



                            For real $y,-1lecos yle1$



                            For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              $$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$



                              For real $y,-1lecos yle1$



                              For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$






                              share|cite|improve this answer









                              $endgroup$



                              $$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$



                              For real $y,-1lecos yle1$



                              For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 18 '18 at 4:43









                              lab bhattacharjeelab bhattacharjee

                              227k15158277




                              227k15158277























                                  0












                                  $begingroup$

                                  By the mean value theorem,



                                  $|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    By the mean value theorem,



                                    $|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      By the mean value theorem,



                                      $|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.






                                      share|cite|improve this answer









                                      $endgroup$



                                      By the mean value theorem,



                                      $|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 18 '18 at 4:53









                                      Mustafa SaidMustafa Said

                                      3,0611913




                                      3,0611913






























                                          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%2f3044759%2fcompute-the-limit-or-show-it-doesnt-exist-lim-n-to-infty-sin-sqrtn1%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

                                          Plaza Victoria

                                          Image
                                          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                                          Read more

                                          Puebla de Zaragoza

                                          Image
                                          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                                          Read more

                                          Musa

                                          Image
                                          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                                          Read more