Determine the limit, or show it doesn't exist: $lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$












4












$begingroup$



Determine the limit of the following or prove it doesn't exist:
$$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$




If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.






calculus limits trigonometry







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$endgroup$

















    4












    $begingroup$



    Determine the limit of the following or prove it doesn't exist:
    $$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$




    If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.






    calculus limits trigonometry







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$



      Determine the limit of the following or prove it doesn't exist:
      $$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$




      If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.






      calculus limits trigonometry







      share|cite|improve this question











      $endgroup$





      Determine the limit of the following or prove it doesn't exist:
      $$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$




      If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.






      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 3:47









      Blue

      47.9k870153




      47.9k870153










      asked Dec 18 '18 at 2:50









      J. LastinJ. Lastin

      986




      986






















          2 Answers
          2






          active

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          5












          $begingroup$

          Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
            $endgroup$
            – J. Lastin
            Dec 18 '18 at 3:06






          • 1




            $begingroup$
            OK I added a bit
            $endgroup$
            – william122
            Dec 18 '18 at 3:10



















          1












          $begingroup$

          Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.



          Just because, let's do it rigorously.

          We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.



          Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$



          Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.






          share|cite|improve this answer









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            2 Answers
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            2 Answers
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            active

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            5












            $begingroup$

            Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
              $endgroup$
              – J. Lastin
              Dec 18 '18 at 3:06






            • 1




              $begingroup$
              OK I added a bit
              $endgroup$
              – william122
              Dec 18 '18 at 3:10
















            5












            $begingroup$

            Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
              $endgroup$
              – J. Lastin
              Dec 18 '18 at 3:06






            • 1




              $begingroup$
              OK I added a bit
              $endgroup$
              – william122
              Dec 18 '18 at 3:10














            5












            5








            5





            $begingroup$

            Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)






            share|cite|improve this answer











            $endgroup$



            Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 3:09

























            answered Dec 18 '18 at 2:58









            william122william122

            52412




            52412












            • $begingroup$
              Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
              $endgroup$
              – J. Lastin
              Dec 18 '18 at 3:06






            • 1




              $begingroup$
              OK I added a bit
              $endgroup$
              – william122
              Dec 18 '18 at 3:10


















            • $begingroup$
              Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
              $endgroup$
              – J. Lastin
              Dec 18 '18 at 3:06






            • 1




              $begingroup$
              OK I added a bit
              $endgroup$
              – william122
              Dec 18 '18 at 3:10
















            $begingroup$
            Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
            $endgroup$
            – J. Lastin
            Dec 18 '18 at 3:06




            $begingroup$
            Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
            $endgroup$
            – J. Lastin
            Dec 18 '18 at 3:06




            1




            1




            $begingroup$
            OK I added a bit
            $endgroup$
            – william122
            Dec 18 '18 at 3:10




            $begingroup$
            OK I added a bit
            $endgroup$
            – william122
            Dec 18 '18 at 3:10











            1












            $begingroup$

            Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.



            Just because, let's do it rigorously.

            We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.



            Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$



            Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.



              Just because, let's do it rigorously.

              We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.



              Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$



              Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.



                Just because, let's do it rigorously.

                We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.



                Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$



                Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.






                share|cite|improve this answer









                $endgroup$



                Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.



                Just because, let's do it rigorously.

                We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.



                Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$



                Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 3:26









                R.JacksonR.Jackson

                1688




                1688






























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