Finding the limit $frac {sin x - arctan x}{x^2 ln x}$ as $x rightarrow 0$ in Petrovic











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Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$



My questions:



1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?



2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?






calculus limits







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  • 1




    Your observations are both correct.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    Yes to both. There are other methods too, such as power series, to show the limit is 0.
    – DanielWainfleet
    2 hours ago















up vote
2
down vote

favorite












Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$



My questions:



1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?



2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?






calculus limits







share|cite|improve this question




















  • 1




    Your observations are both correct.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    Yes to both. There are other methods too, such as power series, to show the limit is 0.
    – DanielWainfleet
    2 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$



My questions:



1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?



2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?






calculus limits







share|cite|improve this question















Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$



My questions:



1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?



2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?






calculus limits




calculus limits






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share|cite|improve this question













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edited 2 hours ago









PradyumanDixit

774214




774214










asked 3 hours ago









hopefully

1868




1868








  • 1




    Your observations are both correct.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    Yes to both. There are other methods too, such as power series, to show the limit is 0.
    – DanielWainfleet
    2 hours ago














  • 1




    Your observations are both correct.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    Yes to both. There are other methods too, such as power series, to show the limit is 0.
    – DanielWainfleet
    2 hours ago








1




1




Your observations are both correct.
– Kavi Rama Murthy
3 hours ago




Your observations are both correct.
– Kavi Rama Murthy
3 hours ago




1




1




Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago




Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago










2 Answers
2






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1
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Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $



Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives



$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$






share|cite|improve this answer








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Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    up vote
    1
    down vote













    Another way:



    $$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$



    Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.



    $$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $



      Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives



      $$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$






      share|cite|improve this answer








      New contributor




      Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















        up vote
        1
        down vote



        accepted










        Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $



        Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives



        $$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$






        share|cite|improve this answer








        New contributor




        Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.




















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $



          Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives



          $$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$






          share|cite|improve this answer








          New contributor




          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $



          Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives



          $$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$







          share|cite|improve this answer








          New contributor




          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 hours ago









          Lakshya Sinha

          364




          364




          New contributor




          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






















              up vote
              1
              down vote













              Another way:



              $$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$



              Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.



              $$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Another way:



                $$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$



                Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.



                $$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Another way:



                  $$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$



                  Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.



                  $$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$






                  share|cite|improve this answer












                  Another way:



                  $$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$



                  Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.



                  $$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  lab bhattacharjee

                  219k14153268




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