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












3














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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    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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    New contributor




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      3












      3








      3


      1





      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









      New contributor




      Heryon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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




      logarithms limits-without-lhopital indeterminate-forms






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      edited yesterday









      mrtaurho

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      asked yesterday









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          4 Answers
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          3














          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





















          • Clever, thanks. Any way to do this without Taylor series? (Just to know)
            – Heryon
            yesterday



















          2














          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,
          $$lim_{xto0}Bigg(frac{d}{dx}Bigg(frac{ln(1+x)}{x}Bigg)Bigg)=lim_{xto0}bigg(-frac{1}{2}+frac{2x}{3}-frac{3x^2}{4}+...bigg)=fbox{$-frac{1}{2}$}$$






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            1














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



            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























            • Yes I can't use that too, edited the post, but thanks!
              – Heryon
              yesterday



















            0














            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





















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              4 Answers
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              active

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              4 Answers
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              active

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              3














              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





















              • Clever, thanks. Any way to do this without Taylor series? (Just to know)
                – Heryon
                yesterday
















              3














              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





















              • Clever, thanks. Any way to do this without Taylor series? (Just to know)
                – Heryon
                yesterday














              3












              3








              3






              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












              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 yesterday









              Aditya Dua

              86418




              86418












              • Clever, thanks. Any way to do this without Taylor series? (Just to know)
                – Heryon
                yesterday


















              • Clever, thanks. Any way to do this without Taylor series? (Just to know)
                – Heryon
                yesterday
















              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              – Heryon
              yesterday




              Clever, thanks. Any way to do this without Taylor series? (Just to know)
              – Heryon
              yesterday











              2














              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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                2














                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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                  2












                  2








                  2






                  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








                  New contributor




                  Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  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








                  New contributor




                  Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




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                  answered yesterday









                  Peter Foreman

                  2056




                  2056




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                      1














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



                      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























                      • Yes I can't use that too, edited the post, but thanks!
                        – Heryon
                        yesterday
















                      1














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



                      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























                      • Yes I can't use that too, edited the post, but thanks!
                        – Heryon
                        yesterday














                      1












                      1








                      1






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



                      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














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



                      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 yesterday

























                      answered yesterday









                      Doug M

                      44.2k31854




                      44.2k31854












                      • Yes I can't use that too, edited the post, but thanks!
                        – Heryon
                        yesterday


















                      • Yes I can't use that too, edited the post, but thanks!
                        – Heryon
                        yesterday
















                      Yes I can't use that too, edited the post, but thanks!
                      – Heryon
                      yesterday




                      Yes I can't use that too, edited the post, but thanks!
                      – Heryon
                      yesterday











                      0














                      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


























                        0














                        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
























                          0












                          0








                          0






                          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












                          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 yesterday









                          李子镔

                          314




                          314






















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