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







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  • 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






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edited Dec 18 '18 at 4:24









tatan

5,63362655




5,63362655










asked Dec 18 '18 at 4:02









J. LastinJ. Lastin

986




986








  • 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









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    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









        $endgroup$













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          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 $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$








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              share|cite|improve this answer










              answered Dec 18 '18 at 4:18









              TravisTravis

              60k767146




              60k767146























                  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









                      l''''''''ll''''''''l

                      2,198726




                      2,198726























                          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

                              224k15156274




                              224k15156274























                                  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

                                      2,9511913




                                      2,9511913






























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