How to find the limit of following sequences (natural logarithm)? [closed]












-1














$lim_{xrightarrowinfty} ::ln x-x$



$lim_{xrightarrow0^+}::ln x-x$






limits logarithms







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closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
    – Ross Millikan
    Dec 28 '18 at 14:10










  • How exactly would those limits be confusing?
    – KM101
    Dec 28 '18 at 14:10






  • 1




    Both limites are $$-infty$$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 14:22
















-1














$lim_{xrightarrowinfty} ::ln x-x$



$lim_{xrightarrow0^+}::ln x-x$






limits logarithms







share|cite|improve this question















closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
    – Ross Millikan
    Dec 28 '18 at 14:10










  • How exactly would those limits be confusing?
    – KM101
    Dec 28 '18 at 14:10






  • 1




    Both limites are $$-infty$$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 14:22














-1












-1








-1







$lim_{xrightarrowinfty} ::ln x-x$



$lim_{xrightarrow0^+}::ln x-x$






limits logarithms







share|cite|improve this question















$lim_{xrightarrowinfty} ::ln x-x$



$lim_{xrightarrow0^+}::ln x-x$






limits logarithms




limits logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 14:24









Rakibul Islam Prince

1,035211




1,035211










asked Dec 28 '18 at 14:02









EmiliaEmilia

31




31




closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
    – Ross Millikan
    Dec 28 '18 at 14:10










  • How exactly would those limits be confusing?
    – KM101
    Dec 28 '18 at 14:10






  • 1




    Both limites are $$-infty$$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 14:22














  • 1




    What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
    – Ross Millikan
    Dec 28 '18 at 14:10










  • How exactly would those limits be confusing?
    – KM101
    Dec 28 '18 at 14:10






  • 1




    Both limites are $$-infty$$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 14:22








1




1




What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10




What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10












How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10




How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10




1




1




Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22




Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22










2 Answers
2






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oldest

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0














You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.






share|cite|improve this answer





























    0














    if $x>0$,we can expand as,
    $$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
    now,
    $$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
    $$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
    $$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$



    now,if
    $$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
    $$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
    $$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
    $$=-infty$$






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      You have that
      $$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
      using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.






      share|cite|improve this answer


























        0














        You have that
        $$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
        using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.






        share|cite|improve this answer
























          0












          0








          0






          You have that
          $$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
          using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.






          share|cite|improve this answer












          You have that
          $$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
          using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 14:23









          José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

          808110




          808110























              0














              if $x>0$,we can expand as,
              $$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
              now,
              $$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
              $$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
              $$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$



              now,if
              $$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
              $$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
              $$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
              $$=-infty$$






              share|cite|improve this answer


























                0














                if $x>0$,we can expand as,
                $$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                now,
                $$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
                $$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
                $$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$



                now,if
                $$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                $$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
                $$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
                $$=-infty$$






                share|cite|improve this answer
























                  0












                  0








                  0






                  if $x>0$,we can expand as,
                  $$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                  now,
                  $$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
                  $$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
                  $$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$



                  now,if
                  $$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                  $$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
                  $$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
                  $$=-infty$$






                  share|cite|improve this answer












                  if $x>0$,we can expand as,
                  $$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                  now,
                  $$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
                  $$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
                  $$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$



                  now,if
                  $$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
                  $$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
                  $$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
                  $$=-infty$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 28 '18 at 14:47









                  Rakibul Islam PrinceRakibul Islam Prince

                  1,035211




                  1,035211















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