Does the limit of $sin (pi/x)$ converge or diverge?












0












$begingroup$


Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$



I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
    $endgroup$
    – Michael Lee
    Dec 2 '18 at 21:36












  • $begingroup$
    You can use an alternating sequence
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:37










  • $begingroup$
    It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 2:04
















0












$begingroup$


Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$



I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
    $endgroup$
    – Michael Lee
    Dec 2 '18 at 21:36












  • $begingroup$
    You can use an alternating sequence
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:37










  • $begingroup$
    It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 2:04














0












0








0





$begingroup$


Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$



I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$




Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$



I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 21:33









Key Flex

7,80961232




7,80961232










asked Dec 2 '18 at 21:31









Ty JohnsonTy Johnson

323




323












  • $begingroup$
    First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
    $endgroup$
    – Michael Lee
    Dec 2 '18 at 21:36












  • $begingroup$
    You can use an alternating sequence
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:37










  • $begingroup$
    It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 2:04


















  • $begingroup$
    First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
    $endgroup$
    – Michael Lee
    Dec 2 '18 at 21:36












  • $begingroup$
    You can use an alternating sequence
    $endgroup$
    – Fakemistake
    Dec 2 '18 at 21:37










  • $begingroup$
    It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 2:04
















$begingroup$
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
$endgroup$
– Michael Lee
Dec 2 '18 at 21:36






$begingroup$
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
$endgroup$
– Michael Lee
Dec 2 '18 at 21:36














$begingroup$
You can use an alternating sequence
$endgroup$
– Fakemistake
Dec 2 '18 at 21:37




$begingroup$
You can use an alternating sequence
$endgroup$
– Fakemistake
Dec 2 '18 at 21:37












$begingroup$
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
$endgroup$
– Paramanand Singh
Dec 3 '18 at 2:04




$begingroup$
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
$endgroup$
– Paramanand Singh
Dec 3 '18 at 2:04










3 Answers
3






active

oldest

votes


















1












$begingroup$

hint



$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$



$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let consider the case $x=frac1yto 0^+ quad y to infty$ then



    $$sin (pi/x)=sin (ypi )$$



    and then the subsequence



    $$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?



      You should try to take a look on the graph of the function.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $x_n$ and $y_n$ are constants?
        $endgroup$
        – Ty Johnson
        Dec 2 '18 at 22:25










      • $begingroup$
        I am sorry, the fuction applied at those points is constanr
        $endgroup$
        – Célio Augusto
        Dec 4 '18 at 22:18











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      hint



      $$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$



      $$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        hint



        $$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$



        $$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          hint



          $$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$



          $$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$






          share|cite|improve this answer









          $endgroup$



          hint



          $$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$



          $$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 21:37









          hamam_Abdallahhamam_Abdallah

          38k21634




          38k21634























              0












              $begingroup$

              Let consider the case $x=frac1yto 0^+ quad y to infty$ then



              $$sin (pi/x)=sin (ypi )$$



              and then the subsequence



              $$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let consider the case $x=frac1yto 0^+ quad y to infty$ then



                $$sin (pi/x)=sin (ypi )$$



                and then the subsequence



                $$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let consider the case $x=frac1yto 0^+ quad y to infty$ then



                  $$sin (pi/x)=sin (ypi )$$



                  and then the subsequence



                  $$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$






                  share|cite|improve this answer









                  $endgroup$



                  Let consider the case $x=frac1yto 0^+ quad y to infty$ then



                  $$sin (pi/x)=sin (ypi )$$



                  and then the subsequence



                  $$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 21:59









                  gimusigimusi

                  92.8k84494




                  92.8k84494























                      0












                      $begingroup$

                      Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?



                      You should try to take a look on the graph of the function.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $x_n$ and $y_n$ are constants?
                        $endgroup$
                        – Ty Johnson
                        Dec 2 '18 at 22:25










                      • $begingroup$
                        I am sorry, the fuction applied at those points is constanr
                        $endgroup$
                        – Célio Augusto
                        Dec 4 '18 at 22:18
















                      0












                      $begingroup$

                      Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?



                      You should try to take a look on the graph of the function.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $x_n$ and $y_n$ are constants?
                        $endgroup$
                        – Ty Johnson
                        Dec 2 '18 at 22:25










                      • $begingroup$
                        I am sorry, the fuction applied at those points is constanr
                        $endgroup$
                        – Célio Augusto
                        Dec 4 '18 at 22:18














                      0












                      0








                      0





                      $begingroup$

                      Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?



                      You should try to take a look on the graph of the function.






                      share|cite|improve this answer











                      $endgroup$



                      Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?



                      You should try to take a look on the graph of the function.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 4 '18 at 22:19

























                      answered Dec 2 '18 at 21:39









                      Célio AugustoCélio Augusto

                      192




                      192












                      • $begingroup$
                        $x_n$ and $y_n$ are constants?
                        $endgroup$
                        – Ty Johnson
                        Dec 2 '18 at 22:25










                      • $begingroup$
                        I am sorry, the fuction applied at those points is constanr
                        $endgroup$
                        – Célio Augusto
                        Dec 4 '18 at 22:18


















                      • $begingroup$
                        $x_n$ and $y_n$ are constants?
                        $endgroup$
                        – Ty Johnson
                        Dec 2 '18 at 22:25










                      • $begingroup$
                        I am sorry, the fuction applied at those points is constanr
                        $endgroup$
                        – Célio Augusto
                        Dec 4 '18 at 22:18
















                      $begingroup$
                      $x_n$ and $y_n$ are constants?
                      $endgroup$
                      – Ty Johnson
                      Dec 2 '18 at 22:25




                      $begingroup$
                      $x_n$ and $y_n$ are constants?
                      $endgroup$
                      – Ty Johnson
                      Dec 2 '18 at 22:25












                      $begingroup$
                      I am sorry, the fuction applied at those points is constanr
                      $endgroup$
                      – Célio Augusto
                      Dec 4 '18 at 22:18




                      $begingroup$
                      I am sorry, the fuction applied at those points is constanr
                      $endgroup$
                      – Célio Augusto
                      Dec 4 '18 at 22:18


















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