prove that Limit of $sinbar{z} over{sinz} $ does not exists as $zto 0$












0












$begingroup$


so i have to prove that the limit of this function doenot exists



$sinbar{z} over{sinz} $ as $zto 0$



well i started with z = $re^{itheta}$



so



${sinspace rspace e^{-itheta}} over sin space r space e^{itheta}$



so $lim z space spacespace when spacespace zto0 $



$ sin space e^{-2itheta} space spacespacespace is space depends space on space theta space only space so space lim space doesn't space exists
space forspace this space function space right space ?? $






calculus abstract-algebra general-topology complex-analysis







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$endgroup$












  • $begingroup$
    but how you get $sin e^{-i2theta}$ in your question? It is not clear
    $endgroup$
    – Masacroso
    Nov 30 '18 at 17:28










  • $begingroup$
    i tried to sole it and that is what i got
    $endgroup$
    – Smb Youz
    Nov 30 '18 at 17:31
















0












$begingroup$


so i have to prove that the limit of this function doenot exists



$sinbar{z} over{sinz} $ as $zto 0$



well i started with z = $re^{itheta}$



so



${sinspace rspace e^{-itheta}} over sin space r space e^{itheta}$



so $lim z space spacespace when spacespace zto0 $



$ sin space e^{-2itheta} space spacespacespace is space depends space on space theta space only space so space lim space doesn't space exists
space forspace this space function space right space ?? $






calculus abstract-algebra general-topology complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    but how you get $sin e^{-i2theta}$ in your question? It is not clear
    $endgroup$
    – Masacroso
    Nov 30 '18 at 17:28










  • $begingroup$
    i tried to sole it and that is what i got
    $endgroup$
    – Smb Youz
    Nov 30 '18 at 17:31














0












0








0





$begingroup$


so i have to prove that the limit of this function doenot exists



$sinbar{z} over{sinz} $ as $zto 0$



well i started with z = $re^{itheta}$



so



${sinspace rspace e^{-itheta}} over sin space r space e^{itheta}$



so $lim z space spacespace when spacespace zto0 $



$ sin space e^{-2itheta} space spacespacespace is space depends space on space theta space only space so space lim space doesn't space exists
space forspace this space function space right space ?? $






calculus abstract-algebra general-topology complex-analysis







share|cite|improve this question











$endgroup$




so i have to prove that the limit of this function doenot exists



$sinbar{z} over{sinz} $ as $zto 0$



well i started with z = $re^{itheta}$



so



${sinspace rspace e^{-itheta}} over sin space r space e^{itheta}$



so $lim z space spacespace when spacespace zto0 $



$ sin space e^{-2itheta} space spacespacespace is space depends space on space theta space only space so space lim space doesn't space exists
space forspace this space function space right space ?? $






calculus abstract-algebra general-topology complex-analysis




calculus abstract-algebra general-topology complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 17:11







Smb Youz

















asked Nov 30 '18 at 17:03









Smb YouzSmb Youz

165




165












  • $begingroup$
    but how you get $sin e^{-i2theta}$ in your question? It is not clear
    $endgroup$
    – Masacroso
    Nov 30 '18 at 17:28










  • $begingroup$
    i tried to sole it and that is what i got
    $endgroup$
    – Smb Youz
    Nov 30 '18 at 17:31


















  • $begingroup$
    but how you get $sin e^{-i2theta}$ in your question? It is not clear
    $endgroup$
    – Masacroso
    Nov 30 '18 at 17:28










  • $begingroup$
    i tried to sole it and that is what i got
    $endgroup$
    – Smb Youz
    Nov 30 '18 at 17:31
















$begingroup$
but how you get $sin e^{-i2theta}$ in your question? It is not clear
$endgroup$
– Masacroso
Nov 30 '18 at 17:28




$begingroup$
but how you get $sin e^{-i2theta}$ in your question? It is not clear
$endgroup$
– Masacroso
Nov 30 '18 at 17:28












$begingroup$
i tried to sole it and that is what i got
$endgroup$
– Smb Youz
Nov 30 '18 at 17:31




$begingroup$
i tried to sole it and that is what i got
$endgroup$
– Smb Youz
Nov 30 '18 at 17:31










2 Answers
2






active

oldest

votes


















2












$begingroup$

If $z = x in mathbb R$, then
$$
frac{sin overline z}{sin z} = frac {sin x}{sin x} = 1 to 1;
$$
if $z = mathrm i y, y in mathbb R$, then
$$
frac {sin overline z}{sin z} = frac {sin(-mathrm i y)}{sin (mathrm i y)},
$$
using the series expansion of $sin z$ we have $sin (-z) = -sin(z)$, so the quotient above equals $-1$.



Thus the limit does not exist.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    From $lim_{zto 0}frac{z}{sin z}=1$ we have that



    $$lim_{zto 0}frac{sinbar z}{sin z}=lim_{zto 0}frac{bar z}{z}=lim_{rto 0^+}frac{re^{-itheta}}{r e^{itheta}}=e^{-i2theta}$$



    for $z=re^{itheta}$. Hence the limit doesn't exists because for different $thetain [0,pi)$ the value $e^{-i2theta}$ changes.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      if the professor ask me in the question to ""show him how"" is that it has no limit !!
      $endgroup$
      – Smb Youz
      Nov 30 '18 at 17:23











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    If $z = x in mathbb R$, then
    $$
    frac{sin overline z}{sin z} = frac {sin x}{sin x} = 1 to 1;
    $$
    if $z = mathrm i y, y in mathbb R$, then
    $$
    frac {sin overline z}{sin z} = frac {sin(-mathrm i y)}{sin (mathrm i y)},
    $$
    using the series expansion of $sin z$ we have $sin (-z) = -sin(z)$, so the quotient above equals $-1$.



    Thus the limit does not exist.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If $z = x in mathbb R$, then
      $$
      frac{sin overline z}{sin z} = frac {sin x}{sin x} = 1 to 1;
      $$
      if $z = mathrm i y, y in mathbb R$, then
      $$
      frac {sin overline z}{sin z} = frac {sin(-mathrm i y)}{sin (mathrm i y)},
      $$
      using the series expansion of $sin z$ we have $sin (-z) = -sin(z)$, so the quotient above equals $-1$.



      Thus the limit does not exist.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        If $z = x in mathbb R$, then
        $$
        frac{sin overline z}{sin z} = frac {sin x}{sin x} = 1 to 1;
        $$
        if $z = mathrm i y, y in mathbb R$, then
        $$
        frac {sin overline z}{sin z} = frac {sin(-mathrm i y)}{sin (mathrm i y)},
        $$
        using the series expansion of $sin z$ we have $sin (-z) = -sin(z)$, so the quotient above equals $-1$.



        Thus the limit does not exist.






        share|cite|improve this answer









        $endgroup$



        If $z = x in mathbb R$, then
        $$
        frac{sin overline z}{sin z} = frac {sin x}{sin x} = 1 to 1;
        $$
        if $z = mathrm i y, y in mathbb R$, then
        $$
        frac {sin overline z}{sin z} = frac {sin(-mathrm i y)}{sin (mathrm i y)},
        $$
        using the series expansion of $sin z$ we have $sin (-z) = -sin(z)$, so the quotient above equals $-1$.



        Thus the limit does not exist.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 17:21









        xbhxbh

        6,1351522




        6,1351522























            1












            $begingroup$

            From $lim_{zto 0}frac{z}{sin z}=1$ we have that



            $$lim_{zto 0}frac{sinbar z}{sin z}=lim_{zto 0}frac{bar z}{z}=lim_{rto 0^+}frac{re^{-itheta}}{r e^{itheta}}=e^{-i2theta}$$



            for $z=re^{itheta}$. Hence the limit doesn't exists because for different $thetain [0,pi)$ the value $e^{-i2theta}$ changes.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if the professor ask me in the question to ""show him how"" is that it has no limit !!
              $endgroup$
              – Smb Youz
              Nov 30 '18 at 17:23
















            1












            $begingroup$

            From $lim_{zto 0}frac{z}{sin z}=1$ we have that



            $$lim_{zto 0}frac{sinbar z}{sin z}=lim_{zto 0}frac{bar z}{z}=lim_{rto 0^+}frac{re^{-itheta}}{r e^{itheta}}=e^{-i2theta}$$



            for $z=re^{itheta}$. Hence the limit doesn't exists because for different $thetain [0,pi)$ the value $e^{-i2theta}$ changes.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if the professor ask me in the question to ""show him how"" is that it has no limit !!
              $endgroup$
              – Smb Youz
              Nov 30 '18 at 17:23














            1












            1








            1





            $begingroup$

            From $lim_{zto 0}frac{z}{sin z}=1$ we have that



            $$lim_{zto 0}frac{sinbar z}{sin z}=lim_{zto 0}frac{bar z}{z}=lim_{rto 0^+}frac{re^{-itheta}}{r e^{itheta}}=e^{-i2theta}$$



            for $z=re^{itheta}$. Hence the limit doesn't exists because for different $thetain [0,pi)$ the value $e^{-i2theta}$ changes.






            share|cite|improve this answer











            $endgroup$



            From $lim_{zto 0}frac{z}{sin z}=1$ we have that



            $$lim_{zto 0}frac{sinbar z}{sin z}=lim_{zto 0}frac{bar z}{z}=lim_{rto 0^+}frac{re^{-itheta}}{r e^{itheta}}=e^{-i2theta}$$



            for $z=re^{itheta}$. Hence the limit doesn't exists because for different $thetain [0,pi)$ the value $e^{-i2theta}$ changes.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 30 '18 at 21:23

























            answered Nov 30 '18 at 17:15









            MasacrosoMasacroso

            13.1k41747




            13.1k41747












            • $begingroup$
              if the professor ask me in the question to ""show him how"" is that it has no limit !!
              $endgroup$
              – Smb Youz
              Nov 30 '18 at 17:23


















            • $begingroup$
              if the professor ask me in the question to ""show him how"" is that it has no limit !!
              $endgroup$
              – Smb Youz
              Nov 30 '18 at 17:23
















            $begingroup$
            if the professor ask me in the question to ""show him how"" is that it has no limit !!
            $endgroup$
            – Smb Youz
            Nov 30 '18 at 17:23




            $begingroup$
            if the professor ask me in the question to ""show him how"" is that it has no limit !!
            $endgroup$
            – Smb Youz
            Nov 30 '18 at 17:23


















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