Calcul of $lim_{xto 1^{-}}frac{pi-arccos(x)}{sqrt{1-x^2}}$












2












$begingroup$


I found that
$$lim_{xto 1^{-}}frac{pi-arccos(x)}{sqrt{1-x^2}}=+infty$$
My question: Can we use Taylor series method to find this limit?






limits taylor-expansion







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I found that
    $$lim_{xto 1^{-}}frac{pi-arccos(x)}{sqrt{1-x^2}}=+infty$$
    My question: Can we use Taylor series method to find this limit?






    limits taylor-expansion







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I found that
      $$lim_{xto 1^{-}}frac{pi-arccos(x)}{sqrt{1-x^2}}=+infty$$
      My question: Can we use Taylor series method to find this limit?






      limits taylor-expansion







      share|cite|improve this question











      $endgroup$




      I found that
      $$lim_{xto 1^{-}}frac{pi-arccos(x)}{sqrt{1-x^2}}=+infty$$
      My question: Can we use Taylor series method to find this limit?






      limits taylor-expansion




      limits taylor-expansion






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 20:25







      Theory Nombre

















      asked Dec 6 '18 at 20:19









      Theory NombreTheory Nombre

      1297




      1297






















          2 Answers
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          $begingroup$

          Note that we cannot use l’Hopital since as $xto1^-$



          $$frac{pi-arccos(x)}{sqrt{1-x^2}}$$



          is in the form $frac{pi}{0^+}$, and we can directly conclude from here that the limit is $+infty$.



          A more interesting limit would be the same for $xto -1^+$ and in that case we can of course apply l'Hopital.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            We can but it is completely not necessary here.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:28






          • 1




            $begingroup$
            The proof is already there ... "We can directly..."
            $endgroup$
            – GEdgar
            Dec 6 '18 at 20:31






          • 1




            $begingroup$
            @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:42






          • 1




            $begingroup$
            @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:44






          • 1




            $begingroup$
            @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:45



















          2












          $begingroup$

          You cannot use Taylor series here, because $arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.



          There is a generalisation you can use though, the so called Puisseux Series.



          As mentioned in the other answer, the more interesting limit is $xto -1^+$.
          As the first right derivative of $arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:



          $$lim_{xto -1^{+}}frac{pi-arccos(x)}{sqrt{1-x^2}}=lim_{xto -1^{+}} frac {-(1-x^2)^{-1/2}}{-x (1-x^2)^{-1/2} } = lim_{xto -1^{+}}frac 1 x =-1.$$



          You could also try to use the mentioned puisseux series: However,
          $$pi -arccos(x) =sqrt2 sqrt{x+1} + frac {(x+1)^{frac 3 2}}{6sqrt 2}+ dots$$
          and $$ sqrt{1-x^2}= sqrt 2 sqrt{x+1} - frac {(x+1)^{frac 3 2}}{2sqrt 2}+dots$$
          and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            2












            $begingroup$

            Note that we cannot use l’Hopital since as $xto1^-$



            $$frac{pi-arccos(x)}{sqrt{1-x^2}}$$



            is in the form $frac{pi}{0^+}$, and we can directly conclude from here that the limit is $+infty$.



            A more interesting limit would be the same for $xto -1^+$ and in that case we can of course apply l'Hopital.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              We can but it is completely not necessary here.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:28






            • 1




              $begingroup$
              The proof is already there ... "We can directly..."
              $endgroup$
              – GEdgar
              Dec 6 '18 at 20:31






            • 1




              $begingroup$
              @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:42






            • 1




              $begingroup$
              @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:44






            • 1




              $begingroup$
              @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:45
















            2












            $begingroup$

            Note that we cannot use l’Hopital since as $xto1^-$



            $$frac{pi-arccos(x)}{sqrt{1-x^2}}$$



            is in the form $frac{pi}{0^+}$, and we can directly conclude from here that the limit is $+infty$.



            A more interesting limit would be the same for $xto -1^+$ and in that case we can of course apply l'Hopital.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              We can but it is completely not necessary here.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:28






            • 1




              $begingroup$
              The proof is already there ... "We can directly..."
              $endgroup$
              – GEdgar
              Dec 6 '18 at 20:31






            • 1




              $begingroup$
              @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:42






            • 1




              $begingroup$
              @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:44






            • 1




              $begingroup$
              @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:45














            2












            2








            2





            $begingroup$

            Note that we cannot use l’Hopital since as $xto1^-$



            $$frac{pi-arccos(x)}{sqrt{1-x^2}}$$



            is in the form $frac{pi}{0^+}$, and we can directly conclude from here that the limit is $+infty$.



            A more interesting limit would be the same for $xto -1^+$ and in that case we can of course apply l'Hopital.






            share|cite|improve this answer











            $endgroup$



            Note that we cannot use l’Hopital since as $xto1^-$



            $$frac{pi-arccos(x)}{sqrt{1-x^2}}$$



            is in the form $frac{pi}{0^+}$, and we can directly conclude from here that the limit is $+infty$.



            A more interesting limit would be the same for $xto -1^+$ and in that case we can of course apply l'Hopital.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 6 '18 at 20:48

























            answered Dec 6 '18 at 20:23









            gimusigimusi

            92.8k84494




            92.8k84494








            • 1




              $begingroup$
              We can but it is completely not necessary here.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:28






            • 1




              $begingroup$
              The proof is already there ... "We can directly..."
              $endgroup$
              – GEdgar
              Dec 6 '18 at 20:31






            • 1




              $begingroup$
              @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:42






            • 1




              $begingroup$
              @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:44






            • 1




              $begingroup$
              @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:45














            • 1




              $begingroup$
              We can but it is completely not necessary here.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:28






            • 1




              $begingroup$
              The proof is already there ... "We can directly..."
              $endgroup$
              – GEdgar
              Dec 6 '18 at 20:31






            • 1




              $begingroup$
              @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:42






            • 1




              $begingroup$
              @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:44






            • 1




              $begingroup$
              @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
              $endgroup$
              – gimusi
              Dec 6 '18 at 20:45








            1




            1




            $begingroup$
            We can but it is completely not necessary here.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:28




            $begingroup$
            We can but it is completely not necessary here.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:28




            1




            1




            $begingroup$
            The proof is already there ... "We can directly..."
            $endgroup$
            – GEdgar
            Dec 6 '18 at 20:31




            $begingroup$
            The proof is already there ... "We can directly..."
            $endgroup$
            – GEdgar
            Dec 6 '18 at 20:31




            1




            1




            $begingroup$
            @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:42




            $begingroup$
            @TheoryNombre As noticed we don't need in that case, we can easily conclude from the fact that the expression is in the form $frac{pi}{0^+}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:42




            1




            1




            $begingroup$
            @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:44




            $begingroup$
            @TheoryNombre We cannot apply Taylor since (arccos x)' doesn't exist at $x=1$ and neither $sqrt{1-x^2}$.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:44




            1




            1




            $begingroup$
            @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:45




            $begingroup$
            @TheoryNombre A more interesting limit would be the same for $xto -1^+$. In that case we can apply l'Hopital.
            $endgroup$
            – gimusi
            Dec 6 '18 at 20:45











            2












            $begingroup$

            You cannot use Taylor series here, because $arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.



            There is a generalisation you can use though, the so called Puisseux Series.



            As mentioned in the other answer, the more interesting limit is $xto -1^+$.
            As the first right derivative of $arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:



            $$lim_{xto -1^{+}}frac{pi-arccos(x)}{sqrt{1-x^2}}=lim_{xto -1^{+}} frac {-(1-x^2)^{-1/2}}{-x (1-x^2)^{-1/2} } = lim_{xto -1^{+}}frac 1 x =-1.$$



            You could also try to use the mentioned puisseux series: However,
            $$pi -arccos(x) =sqrt2 sqrt{x+1} + frac {(x+1)^{frac 3 2}}{6sqrt 2}+ dots$$
            and $$ sqrt{1-x^2}= sqrt 2 sqrt{x+1} - frac {(x+1)^{frac 3 2}}{2sqrt 2}+dots$$
            and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              You cannot use Taylor series here, because $arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.



              There is a generalisation you can use though, the so called Puisseux Series.



              As mentioned in the other answer, the more interesting limit is $xto -1^+$.
              As the first right derivative of $arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:



              $$lim_{xto -1^{+}}frac{pi-arccos(x)}{sqrt{1-x^2}}=lim_{xto -1^{+}} frac {-(1-x^2)^{-1/2}}{-x (1-x^2)^{-1/2} } = lim_{xto -1^{+}}frac 1 x =-1.$$



              You could also try to use the mentioned puisseux series: However,
              $$pi -arccos(x) =sqrt2 sqrt{x+1} + frac {(x+1)^{frac 3 2}}{6sqrt 2}+ dots$$
              and $$ sqrt{1-x^2}= sqrt 2 sqrt{x+1} - frac {(x+1)^{frac 3 2}}{2sqrt 2}+dots$$
              and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                You cannot use Taylor series here, because $arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.



                There is a generalisation you can use though, the so called Puisseux Series.



                As mentioned in the other answer, the more interesting limit is $xto -1^+$.
                As the first right derivative of $arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:



                $$lim_{xto -1^{+}}frac{pi-arccos(x)}{sqrt{1-x^2}}=lim_{xto -1^{+}} frac {-(1-x^2)^{-1/2}}{-x (1-x^2)^{-1/2} } = lim_{xto -1^{+}}frac 1 x =-1.$$



                You could also try to use the mentioned puisseux series: However,
                $$pi -arccos(x) =sqrt2 sqrt{x+1} + frac {(x+1)^{frac 3 2}}{6sqrt 2}+ dots$$
                and $$ sqrt{1-x^2}= sqrt 2 sqrt{x+1} - frac {(x+1)^{frac 3 2}}{2sqrt 2}+dots$$
                and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.






                share|cite|improve this answer











                $endgroup$



                You cannot use Taylor series here, because $arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.



                There is a generalisation you can use though, the so called Puisseux Series.



                As mentioned in the other answer, the more interesting limit is $xto -1^+$.
                As the first right derivative of $arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:



                $$lim_{xto -1^{+}}frac{pi-arccos(x)}{sqrt{1-x^2}}=lim_{xto -1^{+}} frac {-(1-x^2)^{-1/2}}{-x (1-x^2)^{-1/2} } = lim_{xto -1^{+}}frac 1 x =-1.$$



                You could also try to use the mentioned puisseux series: However,
                $$pi -arccos(x) =sqrt2 sqrt{x+1} + frac {(x+1)^{frac 3 2}}{6sqrt 2}+ dots$$
                and $$ sqrt{1-x^2}= sqrt 2 sqrt{x+1} - frac {(x+1)^{frac 3 2}}{2sqrt 2}+dots$$
                and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 21:20

























                answered Dec 6 '18 at 20:40









                klirkklirk

                2,631530




                2,631530






























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