Limlit of $frac{x-xln(1+x)-ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series












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I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






logarithms limits-without-lhopital indeterminate-forms







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    $begingroup$


    I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
    I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






    logarithms limits-without-lhopital indeterminate-forms







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
      I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






      logarithms limits-without-lhopital indeterminate-forms







      share|cite|improve this question











      $endgroup$




      I have $f : x mapsto frac{ln(1+x)}{x}$ which derivative is $$frac{1}{x(1+x)}-frac{ln(1+x)}{x^2}$$
      I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$frac{x - xln(1+x) - ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)






      logarithms limits-without-lhopital indeterminate-forms




      logarithms limits-without-lhopital indeterminate-forms






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      edited Jan 5 at 0:58









      mrtaurho

      6,11271641




      6,11271641










      asked Jan 4 at 23:41









      HeryonHeryon

      214




      214






















          4 Answers
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          $begingroup$

          First, you can decompose the first term in your derivative as:



          $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



          Second, you can expand the second term using Taylor series:



          $frac{ln(1+x)}{x^2}$



          $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



          $ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $



          Putting it together, your derivative is:



          $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




          • The $1/x$ terms cancel out.


          • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


          • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


          So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Clever, thanks. Any way to do this without Taylor series? (Just to know)
            $endgroup$
            – Heryon
            Jan 5 at 0:06



















          2












          $begingroup$

          Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
          $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
          $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
          $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
          So,
          $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






          share|cite|improve this answer









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            1












            $begingroup$

            $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



            At which point I am inclined to use a series



            $frac {d}{dx} (1 - frac x2 + cdots)$



            as $x$ goes to $0$ equals $-frac 12$



            Without using a series, I have.



            $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



            is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes I can't use that too, edited the post, but thanks!
              $endgroup$
              – Heryon
              Jan 5 at 0:33



















            0












            $begingroup$

            Because $frac{x}{1+x}<ln(1+x)<x$



            So when $ x rightarrow 0+, f(x) rightarrow 1$



            When $xrightarrow0-$ is similar






            share|cite|improve this answer









            $endgroup$














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              4 Answers
              4






              active

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              4 Answers
              4






              active

              oldest

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              active

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              active

              oldest

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              3












              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



              $ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06
















              3












              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



              $ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06














              3












              3








              3





              $begingroup$

              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



              $ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$



              First, you can decompose the first term in your derivative as:



              $frac{x}{1+x} = frac{1}{x}-frac{1}{1+x}$.



              Second, you can expand the second term using Taylor series:



              $frac{ln(1+x)}{x^2}$



              $ = frac{1}{x^2} left( x - frac{x^2}{2} + frac{x^3}{3} - .. right)$



              $ = frac{1}{x} - frac{1}{2} + frac{x}{3} - $



              Putting it together, your derivative is:



              $frac{1}{x}-frac{1}{1+x} - frac{1}{x} + frac{1}{2} - frac{x}{3} +...$




              • The $1/x$ terms cancel out.


              • $lim_{x rightarrow 0} frac{1}{1+x} = 1$.


              • $lim_{x rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.


              So you are left with $-1 + frac{1}{2} = -frac{1}{2}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 4 at 23:51









              Aditya DuaAditya Dua

              1,15418




              1,15418












              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06


















              • $begingroup$
                Clever, thanks. Any way to do this without Taylor series? (Just to know)
                $endgroup$
                – Heryon
                Jan 5 at 0:06
















              $begingroup$
              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              $endgroup$
              – Heryon
              Jan 5 at 0:06




              $begingroup$
              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              $endgroup$
              – Heryon
              Jan 5 at 0:06











              2












              $begingroup$

              Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
              $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
              $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
              $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
              So,
              $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                So,
                $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                  $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                  $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                  $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                  So,
                  $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






                  share|cite|improve this answer









                  $endgroup$



                  Your original function $f(x)=frac{ln(1+x)}{x}$ can be written using the taylor series expansion of $ln(1+x)$ as follows:
                  $$ln(1+x)=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k}}{k}=x-frac{x^2}{2}+frac{x^3}{3}-...$$
                  $$thereforefrac{ln(1+x)}{x}=sum_{k=1}^{infty}frac{(-1)^{k+1}x^{k-1}}{k}=sum_{k=0}^{infty}frac{(-1)^{k}x^{k}}{k+1}=1-frac{x}{2}+frac{x^2}{3}-...$$
                  $$frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)=frac{d}{dx}bigg(1-frac{x}{2}+frac{x^2}{3}-...bigg)=-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...$$
                  So,
                  $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 0:17









                  Peter ForemanPeter Foreman

                  5,8141317




                  5,8141317























                      1












                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33
















                      1












                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33














                      1












                      1








                      1





                      $begingroup$

                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$






                      share|cite|improve this answer











                      $endgroup$



                      $frac {1}{x(x+1)} - frac {ln(x+1)}{x^2} = frac {d}{dx} frac {ln(x+1)}{x}$



                      At which point I am inclined to use a series



                      $frac {d}{dx} (1 - frac x2 + cdots)$



                      as $x$ goes to $0$ equals $-frac 12$



                      Without using a series, I have.



                      $frac {d}{dx} frac {ln(x+1)}{x}$ evaluated at $x = 0$



                      is $lim_limits{hto 0} frac{ ln(1+h) - 1}{h^2}$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 5 at 1:30

























                      answered Jan 5 at 0:27









                      Doug MDoug M

                      45.3k31954




                      45.3k31954












                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33


















                      • $begingroup$
                        Yes I can't use that too, edited the post, but thanks!
                        $endgroup$
                        – Heryon
                        Jan 5 at 0:33
















                      $begingroup$
                      Yes I can't use that too, edited the post, but thanks!
                      $endgroup$
                      – Heryon
                      Jan 5 at 0:33




                      $begingroup$
                      Yes I can't use that too, edited the post, but thanks!
                      $endgroup$
                      – Heryon
                      Jan 5 at 0:33











                      0












                      $begingroup$

                      Because $frac{x}{1+x}<ln(1+x)<x$



                      So when $ x rightarrow 0+, f(x) rightarrow 1$



                      When $xrightarrow0-$ is similar






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Because $frac{x}{1+x}<ln(1+x)<x$



                        So when $ x rightarrow 0+, f(x) rightarrow 1$



                        When $xrightarrow0-$ is similar






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Because $frac{x}{1+x}<ln(1+x)<x$



                          So when $ x rightarrow 0+, f(x) rightarrow 1$



                          When $xrightarrow0-$ is similar






                          share|cite|improve this answer









                          $endgroup$



                          Because $frac{x}{1+x}<ln(1+x)<x$



                          So when $ x rightarrow 0+, f(x) rightarrow 1$



                          When $xrightarrow0-$ is similar







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 at 1:46









                          李子镔李子镔

                          314




                          314






























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                              How do I know what Microsoft account the skydrive app is syncing to?

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                              up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
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                              When does type information flow backwards in C++?

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                              up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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                              Grease: Live!

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                              A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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