About the type of singularity [on hold]

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$ f(x)$= 1/($sintheta$ - $costheta$) is having singularity of pole or removable singularity.
$theta $ = π/4 is a singularity.



Pole : $lim_theta to π/4 $f(x) = $infty$



but if we calculate it like this
$lim_theta to π/4 $ $1/(sqrt2)$/($1/(sqrt2)$* $sintheta$ - $1/(sqrt2)$* $costheta$)



so this limit exists.



So as per singularity behavior if the limit exists, it has to be a removable singularity and if it's infinity then it's a pole.



In this case what will be this ?




complex-analysis




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put on hold as unclear what you're asking by José Carlos Santos, user 108128, Jyrki Lahtonen, Adrian Keister, Xander Henderson Aug 5 at 1:59


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















    up vote
    -1
    down vote

    favorite












    $ f(x)$= 1/($sintheta$ - $costheta$) is having singularity of pole or removable singularity.
    $theta $ = π/4 is a singularity.



    Pole : $lim_theta to π/4 $f(x) = $infty$



    but if we calculate it like this
    $lim_theta to π/4 $ $1/(sqrt2)$/($1/(sqrt2)$* $sintheta$ - $1/(sqrt2)$* $costheta$)



    so this limit exists.



    So as per singularity behavior if the limit exists, it has to be a removable singularity and if it's infinity then it's a pole.



    In this case what will be this ?




    complex-analysis




    share|cite|improve this question













    put on hold as unclear what you're asking by José Carlos Santos, user 108128, Jyrki Lahtonen, Adrian Keister, Xander Henderson Aug 5 at 1:59


    Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      $ f(x)$= 1/($sintheta$ - $costheta$) is having singularity of pole or removable singularity.
      $theta $ = π/4 is a singularity.



      Pole : $lim_theta to π/4 $f(x) = $infty$



      but if we calculate it like this
      $lim_theta to π/4 $ $1/(sqrt2)$/($1/(sqrt2)$* $sintheta$ - $1/(sqrt2)$* $costheta$)



      so this limit exists.



      So as per singularity behavior if the limit exists, it has to be a removable singularity and if it's infinity then it's a pole.



      In this case what will be this ?




      complex-analysis




      share|cite|improve this question













      $ f(x)$= 1/($sintheta$ - $costheta$) is having singularity of pole or removable singularity.
      $theta $ = π/4 is a singularity.



      Pole : $lim_theta to π/4 $f(x) = $infty$



      but if we calculate it like this
      $lim_theta to π/4 $ $1/(sqrt2)$/($1/(sqrt2)$* $sintheta$ - $1/(sqrt2)$* $costheta$)



      so this limit exists.



      So as per singularity behavior if the limit exists, it has to be a removable singularity and if it's infinity then it's a pole.



      In this case what will be this ?





      complex-analysis





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 31 at 8:17









      Bernard

      110k635102




      110k635102









      asked Jul 31 at 7:00









      ashish agarwal

      22




      22




      put on hold as unclear what you're asking by José Carlos Santos, user 108128, Jyrki Lahtonen, Adrian Keister, Xander Henderson Aug 5 at 1:59


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






      put on hold as unclear what you're asking by José Carlos Santos, user 108128, Jyrki Lahtonen, Adrian Keister, Xander Henderson Aug 5 at 1:59


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






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          Hint:



          Note that
          $$sin x - cos x = sqrt2cdot sin left( x-fracpi4right)$$






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          • yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
            – ashish agarwal
            Jul 31 at 7:52










          • Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
            – trancelocation
            Jul 31 at 7:59

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          Hint:



          Note that
          $$sin x - cos x = sqrt2cdot sin left( x-fracpi4right)$$






          share|cite|improve this answer





















          • yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
            – ashish agarwal
            Jul 31 at 7:52










          • Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
            – trancelocation
            Jul 31 at 7:59














          up vote
          0
          down vote













          Hint:



          Note that
          $$sin x - cos x = sqrt2cdot sin left( x-fracpi4right)$$






          share|cite|improve this answer





















          • yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
            – ashish agarwal
            Jul 31 at 7:52










          • Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
            – trancelocation
            Jul 31 at 7:59












          up vote
          0
          down vote










          up vote
          0
          down vote









          Hint:



          Note that
          $$sin x - cos x = sqrt2cdot sin left( x-fracpi4right)$$






          share|cite|improve this answer













          Hint:



          Note that
          $$sin x - cos x = sqrt2cdot sin left( x-fracpi4right)$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 7:23









          trancelocation

          4,5701413




          4,5701413











          • yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
            – ashish agarwal
            Jul 31 at 7:52










          • Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
            – trancelocation
            Jul 31 at 7:59
















          • yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
            – ashish agarwal
            Jul 31 at 7:52










          • Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
            – trancelocation
            Jul 31 at 7:59















          yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
          – ashish agarwal
          Jul 31 at 7:52




          yes, thats what I am saying there as well. It should be removable singularity as limit exits. ? In my text books its given as pole thats why I was confused about.
          – ashish agarwal
          Jul 31 at 7:52












          Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
          – trancelocation
          Jul 31 at 7:59




          Obviously, for $f(x) = sqrt2cdot sin left( x-fracpi4right)$, you have $f(fracpi4) = 0$ and $f'(fracpi4) = sqrt2cdot cos 0 neq 0$. So, you have got a pole of order 1.
          – trancelocation
          Jul 31 at 7:59


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