How can I deduce that $limlimits_{xto0} frac{ln(1+x)}x=1$ without Taylor series or L'Hospital's rule?












5














How can I calculate this limit without Taylor series and L'Hospital's rule?
$$lim _{xto0} frac{ln(1+x)}x=1$$






calculus limits logarithms limits-without-lhopital







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  • 5




    It mostly depends on how you define the logarithm.
    – egreg
    Jan 3 '16 at 16:05










  • See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
    – Martin Sleziak
    Nov 25 at 18:30
















5














How can I calculate this limit without Taylor series and L'Hospital's rule?
$$lim _{xto0} frac{ln(1+x)}x=1$$






calculus limits logarithms limits-without-lhopital







share|cite|improve this question




















  • 5




    It mostly depends on how you define the logarithm.
    – egreg
    Jan 3 '16 at 16:05










  • See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
    – Martin Sleziak
    Nov 25 at 18:30














5












5








5







How can I calculate this limit without Taylor series and L'Hospital's rule?
$$lim _{xto0} frac{ln(1+x)}x=1$$






calculus limits logarithms limits-without-lhopital







share|cite|improve this question















How can I calculate this limit without Taylor series and L'Hospital's rule?
$$lim _{xto0} frac{ln(1+x)}x=1$$






calculus limits logarithms limits-without-lhopital




calculus limits logarithms limits-without-lhopital






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share|cite|improve this question













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edited Nov 25 at 18:28









Martin Sleziak

44.6k7115270




44.6k7115270










asked Jan 3 '16 at 15:58









Lovro Sindičić

243216




243216








  • 5




    It mostly depends on how you define the logarithm.
    – egreg
    Jan 3 '16 at 16:05










  • See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
    – Martin Sleziak
    Nov 25 at 18:30














  • 5




    It mostly depends on how you define the logarithm.
    – egreg
    Jan 3 '16 at 16:05










  • See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
    – Martin Sleziak
    Nov 25 at 18:30








5




5




It mostly depends on how you define the logarithm.
– egreg
Jan 3 '16 at 16:05




It mostly depends on how you define the logarithm.
– egreg
Jan 3 '16 at 16:05












See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
– Martin Sleziak
Nov 25 at 18:30




See also: Determine the following limit as x approaches 0: $frac{ln(1+x)}x$ (Although that question does not have restriction to avoid Taylor and L'H.)
– Martin Sleziak
Nov 25 at 18:30










4 Answers
4






active

oldest

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7














If the limit exists, say $L$, then you can state that:
$$begin{align}
L&=lim_{xto0}frac{ln(1+x)}{x}\
therefore e^L&=e^{lim_{xto0}frac{ln(1+x)}{x}}\
&=lim_{xto0}e^{frac{ln(1+x)}{x}}\
&=lim_{xto0}(e^{ln(1+x)})^frac{1}{x}\
&=lim_{xto0}(1+x)^frac{1}{x}\
&=e\
therefore L&=1
end{align}$$






share|cite|improve this answer





























    5














    Note $x geq log (1+x) geq frac{x}{1+x} $ for all $x > -1$. Since $frac{x}{x} to 1$ and $frac{frac{x}{1+x}}{x} to 1$ as $x to 0$. So the limit is $1$.






    share|cite|improve this answer































      3














      Change the variable and use the continuity of $ln(x)$



      $$lim_{x to 0} {ln(1+x) over x} = lim_{n to infty} {lnleft(1+{1 over n}right) over {1over n} }=lim_{n to infty} ncdot{lnleft(1+{1 over n}right) }= lim_{n to infty}{left( lnleft(1+{1 over n}right)^n right) } $$ $$ = lnleft( lim_{n to infty}{left( 1+{1 over n}right)^n }right) = ln(e) = 1$$






      share|cite|improve this answer































        2














        Hint :
        It is equal to the caltulation of $lim _{ xrightarrow 0 }{ { left( x+1 right) }^{ frac { 1 }{ x } } } =e$ just substitute $$t={ left( x+1 right) }^{ frac { 1 }{ x } }$$






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7














          If the limit exists, say $L$, then you can state that:
          $$begin{align}
          L&=lim_{xto0}frac{ln(1+x)}{x}\
          therefore e^L&=e^{lim_{xto0}frac{ln(1+x)}{x}}\
          &=lim_{xto0}e^{frac{ln(1+x)}{x}}\
          &=lim_{xto0}(e^{ln(1+x)})^frac{1}{x}\
          &=lim_{xto0}(1+x)^frac{1}{x}\
          &=e\
          therefore L&=1
          end{align}$$






          share|cite|improve this answer


























            7














            If the limit exists, say $L$, then you can state that:
            $$begin{align}
            L&=lim_{xto0}frac{ln(1+x)}{x}\
            therefore e^L&=e^{lim_{xto0}frac{ln(1+x)}{x}}\
            &=lim_{xto0}e^{frac{ln(1+x)}{x}}\
            &=lim_{xto0}(e^{ln(1+x)})^frac{1}{x}\
            &=lim_{xto0}(1+x)^frac{1}{x}\
            &=e\
            therefore L&=1
            end{align}$$






            share|cite|improve this answer
























              7












              7








              7






              If the limit exists, say $L$, then you can state that:
              $$begin{align}
              L&=lim_{xto0}frac{ln(1+x)}{x}\
              therefore e^L&=e^{lim_{xto0}frac{ln(1+x)}{x}}\
              &=lim_{xto0}e^{frac{ln(1+x)}{x}}\
              &=lim_{xto0}(e^{ln(1+x)})^frac{1}{x}\
              &=lim_{xto0}(1+x)^frac{1}{x}\
              &=e\
              therefore L&=1
              end{align}$$






              share|cite|improve this answer












              If the limit exists, say $L$, then you can state that:
              $$begin{align}
              L&=lim_{xto0}frac{ln(1+x)}{x}\
              therefore e^L&=e^{lim_{xto0}frac{ln(1+x)}{x}}\
              &=lim_{xto0}e^{frac{ln(1+x)}{x}}\
              &=lim_{xto0}(e^{ln(1+x)})^frac{1}{x}\
              &=lim_{xto0}(1+x)^frac{1}{x}\
              &=e\
              therefore L&=1
              end{align}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 3 '16 at 16:05









              Mufasa

              5,0181323




              5,0181323























                  5














                  Note $x geq log (1+x) geq frac{x}{1+x} $ for all $x > -1$. Since $frac{x}{x} to 1$ and $frac{frac{x}{1+x}}{x} to 1$ as $x to 0$. So the limit is $1$.






                  share|cite|improve this answer




























                    5














                    Note $x geq log (1+x) geq frac{x}{1+x} $ for all $x > -1$. Since $frac{x}{x} to 1$ and $frac{frac{x}{1+x}}{x} to 1$ as $x to 0$. So the limit is $1$.






                    share|cite|improve this answer


























                      5












                      5








                      5






                      Note $x geq log (1+x) geq frac{x}{1+x} $ for all $x > -1$. Since $frac{x}{x} to 1$ and $frac{frac{x}{1+x}}{x} to 1$ as $x to 0$. So the limit is $1$.






                      share|cite|improve this answer














                      Note $x geq log (1+x) geq frac{x}{1+x} $ for all $x > -1$. Since $frac{x}{x} to 1$ and $frac{frac{x}{1+x}}{x} to 1$ as $x to 0$. So the limit is $1$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 3 '16 at 16:09

























                      answered Jan 3 '16 at 16:04









                      Batman

                      16.3k11534




                      16.3k11534























                          3














                          Change the variable and use the continuity of $ln(x)$



                          $$lim_{x to 0} {ln(1+x) over x} = lim_{n to infty} {lnleft(1+{1 over n}right) over {1over n} }=lim_{n to infty} ncdot{lnleft(1+{1 over n}right) }= lim_{n to infty}{left( lnleft(1+{1 over n}right)^n right) } $$ $$ = lnleft( lim_{n to infty}{left( 1+{1 over n}right)^n }right) = ln(e) = 1$$






                          share|cite|improve this answer




























                            3














                            Change the variable and use the continuity of $ln(x)$



                            $$lim_{x to 0} {ln(1+x) over x} = lim_{n to infty} {lnleft(1+{1 over n}right) over {1over n} }=lim_{n to infty} ncdot{lnleft(1+{1 over n}right) }= lim_{n to infty}{left( lnleft(1+{1 over n}right)^n right) } $$ $$ = lnleft( lim_{n to infty}{left( 1+{1 over n}right)^n }right) = ln(e) = 1$$






                            share|cite|improve this answer


























                              3












                              3








                              3






                              Change the variable and use the continuity of $ln(x)$



                              $$lim_{x to 0} {ln(1+x) over x} = lim_{n to infty} {lnleft(1+{1 over n}right) over {1over n} }=lim_{n to infty} ncdot{lnleft(1+{1 over n}right) }= lim_{n to infty}{left( lnleft(1+{1 over n}right)^n right) } $$ $$ = lnleft( lim_{n to infty}{left( 1+{1 over n}right)^n }right) = ln(e) = 1$$






                              share|cite|improve this answer














                              Change the variable and use the continuity of $ln(x)$



                              $$lim_{x to 0} {ln(1+x) over x} = lim_{n to infty} {lnleft(1+{1 over n}right) over {1over n} }=lim_{n to infty} ncdot{lnleft(1+{1 over n}right) }= lim_{n to infty}{left( lnleft(1+{1 over n}right)^n right) } $$ $$ = lnleft( lim_{n to infty}{left( 1+{1 over n}right)^n }right) = ln(e) = 1$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Feb 18 '16 at 15:44

























                              answered Jan 3 '16 at 16:04









                              Adrian

                              5,1891135




                              5,1891135























                                  2














                                  Hint :
                                  It is equal to the caltulation of $lim _{ xrightarrow 0 }{ { left( x+1 right) }^{ frac { 1 }{ x } } } =e$ just substitute $$t={ left( x+1 right) }^{ frac { 1 }{ x } }$$






                                  share|cite|improve this answer


























                                    2














                                    Hint :
                                    It is equal to the caltulation of $lim _{ xrightarrow 0 }{ { left( x+1 right) }^{ frac { 1 }{ x } } } =e$ just substitute $$t={ left( x+1 right) }^{ frac { 1 }{ x } }$$






                                    share|cite|improve this answer
























                                      2












                                      2








                                      2






                                      Hint :
                                      It is equal to the caltulation of $lim _{ xrightarrow 0 }{ { left( x+1 right) }^{ frac { 1 }{ x } } } =e$ just substitute $$t={ left( x+1 right) }^{ frac { 1 }{ x } }$$






                                      share|cite|improve this answer












                                      Hint :
                                      It is equal to the caltulation of $lim _{ xrightarrow 0 }{ { left( x+1 right) }^{ frac { 1 }{ x } } } =e$ just substitute $$t={ left( x+1 right) }^{ frac { 1 }{ x } }$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 3 '16 at 16:04









                                      haqnatural

                                      20.6k72457




                                      20.6k72457






























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