Proving that $lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$












9














I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.



It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.






calculus sequences-and-series limits exponentiation







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  • What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
    – Thomas Shelby
    Dec 1 at 18:24










  • @ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
    – Riccardo Cazzin
    Dec 1 at 18:26










  • Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
    – Thomas Andrews
    Dec 1 at 18:26
















9














I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.



It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.






calculus sequences-and-series limits exponentiation







share|cite|improve this question
























  • What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
    – Thomas Shelby
    Dec 1 at 18:24










  • @ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
    – Riccardo Cazzin
    Dec 1 at 18:26










  • Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
    – Thomas Andrews
    Dec 1 at 18:26














9












9








9


1





I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.



It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.






calculus sequences-and-series limits exponentiation







share|cite|improve this question















I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.



It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.






calculus sequences-and-series limits exponentiation




calculus sequences-and-series limits exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 2 at 9:28









Asaf Karagila♦

301k32423755




301k32423755










asked Dec 1 at 18:21









Riccardo Cazzin

1905




1905












  • What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
    – Thomas Shelby
    Dec 1 at 18:24










  • @ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
    – Riccardo Cazzin
    Dec 1 at 18:26










  • Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
    – Thomas Andrews
    Dec 1 at 18:26


















  • What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
    – Thomas Shelby
    Dec 1 at 18:24










  • @ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
    – Riccardo Cazzin
    Dec 1 at 18:26










  • Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
    – Thomas Andrews
    Dec 1 at 18:26
















What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24




What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24












@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26




@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26












Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26




Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26










5 Answers
5






active

oldest

votes


















14














We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$






share|cite|improve this answer





























    11














    That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}






    share|cite|improve this answer

















    • 2




      The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
      – Clement C.
      Dec 1 at 18:37












    • @ClementC. Yes, I agree that one should be careful here.
      – José Carlos Santos
      Dec 1 at 18:48






    • 2




      I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
      – user1717828
      Dec 2 at 1:37






    • 1




      @user1717828 Yes.This happens around here. Get used to it.
      – José Carlos Santos
      Dec 2 at 11:09










    • @user1717828 what can you do...
      – Clement C.
      Dec 4 at 22:17



















    5














    You have
    $$
    left(1+frac{1}{f(n)}right)^{g(n)}
    = expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
    $$
    Since $lim_{ntoinfty} f(n) = infty$, we have
    $$
    g(n) ln left(1+frac{1}{f(n)}right)
    = g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
    = frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
    $$
    and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
    (the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).



    Then,
    $$
    lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
    = e^0 = 1.
    $$






    share|cite|improve this answer

















    • 2




      THIS is the rigorous way (+1)
      – Robert Z
      Dec 2 at 17:10





















    3














    Note that



    $$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$






    share|cite|improve this answer





















    • The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
      – Clement C.
      Dec 1 at 18:35





















    0














    Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
    $$
    g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
    $$
    since
    $$
    frac{g(n)}{f(n)}to0
    $$
    as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
    $$
    lim_{xto0}frac{log (1+x)-0}{x-0}=1
    $$
    by definition of the derivative.






    share|cite|improve this answer





















    • So the result is 0*1 ?
      – Samy Bencherif
      Dec 2 at 6:42










    • No to the limit is $exp(0times 1)$
      – Foobaz John
      Dec 2 at 15:57













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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    14














    We can use that
    $$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$






    share|cite|improve this answer


























      14














      We can use that
      $$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$






      share|cite|improve this answer
























        14












        14








        14






        We can use that
        $$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$






        share|cite|improve this answer












        We can use that
        $$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 18:26









        gimusi

        1




        1























            11














            That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}






            share|cite|improve this answer

















            • 2




              The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
              – Clement C.
              Dec 1 at 18:37












            • @ClementC. Yes, I agree that one should be careful here.
              – José Carlos Santos
              Dec 1 at 18:48






            • 2




              I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
              – user1717828
              Dec 2 at 1:37






            • 1




              @user1717828 Yes.This happens around here. Get used to it.
              – José Carlos Santos
              Dec 2 at 11:09










            • @user1717828 what can you do...
              – Clement C.
              Dec 4 at 22:17
















            11














            That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}






            share|cite|improve this answer

















            • 2




              The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
              – Clement C.
              Dec 1 at 18:37












            • @ClementC. Yes, I agree that one should be careful here.
              – José Carlos Santos
              Dec 1 at 18:48






            • 2




              I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
              – user1717828
              Dec 2 at 1:37






            • 1




              @user1717828 Yes.This happens around here. Get used to it.
              – José Carlos Santos
              Dec 2 at 11:09










            • @user1717828 what can you do...
              – Clement C.
              Dec 4 at 22:17














            11












            11








            11






            That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}






            share|cite|improve this answer












            That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 18:27









            José Carlos Santos

            149k22117219




            149k22117219








            • 2




              The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
              – Clement C.
              Dec 1 at 18:37












            • @ClementC. Yes, I agree that one should be careful here.
              – José Carlos Santos
              Dec 1 at 18:48






            • 2




              I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
              – user1717828
              Dec 2 at 1:37






            • 1




              @user1717828 Yes.This happens around here. Get used to it.
              – José Carlos Santos
              Dec 2 at 11:09










            • @user1717828 what can you do...
              – Clement C.
              Dec 4 at 22:17














            • 2




              The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
              – Clement C.
              Dec 1 at 18:37












            • @ClementC. Yes, I agree that one should be careful here.
              – José Carlos Santos
              Dec 1 at 18:48






            • 2




              I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
              – user1717828
              Dec 2 at 1:37






            • 1




              @user1717828 Yes.This happens around here. Get used to it.
              – José Carlos Santos
              Dec 2 at 11:09










            • @user1717828 what can you do...
              – Clement C.
              Dec 4 at 22:17








            2




            2




            The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
            – Clement C.
            Dec 1 at 18:37






            The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
            – Clement C.
            Dec 1 at 18:37














            @ClementC. Yes, I agree that one should be careful here.
            – José Carlos Santos
            Dec 1 at 18:48




            @ClementC. Yes, I agree that one should be careful here.
            – José Carlos Santos
            Dec 1 at 18:48




            2




            2




            I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
            – user1717828
            Dec 2 at 1:37




            I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
            – user1717828
            Dec 2 at 1:37




            1




            1




            @user1717828 Yes.This happens around here. Get used to it.
            – José Carlos Santos
            Dec 2 at 11:09




            @user1717828 Yes.This happens around here. Get used to it.
            – José Carlos Santos
            Dec 2 at 11:09












            @user1717828 what can you do...
            – Clement C.
            Dec 4 at 22:17




            @user1717828 what can you do...
            – Clement C.
            Dec 4 at 22:17











            5














            You have
            $$
            left(1+frac{1}{f(n)}right)^{g(n)}
            = expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
            $$
            Since $lim_{ntoinfty} f(n) = infty$, we have
            $$
            g(n) ln left(1+frac{1}{f(n)}right)
            = g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
            = frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
            $$
            and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
            (the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).



            Then,
            $$
            lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
            = e^0 = 1.
            $$






            share|cite|improve this answer

















            • 2




              THIS is the rigorous way (+1)
              – Robert Z
              Dec 2 at 17:10


















            5














            You have
            $$
            left(1+frac{1}{f(n)}right)^{g(n)}
            = expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
            $$
            Since $lim_{ntoinfty} f(n) = infty$, we have
            $$
            g(n) ln left(1+frac{1}{f(n)}right)
            = g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
            = frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
            $$
            and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
            (the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).



            Then,
            $$
            lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
            = e^0 = 1.
            $$






            share|cite|improve this answer

















            • 2




              THIS is the rigorous way (+1)
              – Robert Z
              Dec 2 at 17:10
















            5












            5








            5






            You have
            $$
            left(1+frac{1}{f(n)}right)^{g(n)}
            = expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
            $$
            Since $lim_{ntoinfty} f(n) = infty$, we have
            $$
            g(n) ln left(1+frac{1}{f(n)}right)
            = g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
            = frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
            $$
            and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
            (the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).



            Then,
            $$
            lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
            = e^0 = 1.
            $$






            share|cite|improve this answer












            You have
            $$
            left(1+frac{1}{f(n)}right)^{g(n)}
            = expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
            $$
            Since $lim_{ntoinfty} f(n) = infty$, we have
            $$
            g(n) ln left(1+frac{1}{f(n)}right)
            = g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
            = frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
            $$
            and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
            (the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).



            Then,
            $$
            lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
            = e^0 = 1.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 18:32









            Clement C.

            49.6k33886




            49.6k33886








            • 2




              THIS is the rigorous way (+1)
              – Robert Z
              Dec 2 at 17:10
















            • 2




              THIS is the rigorous way (+1)
              – Robert Z
              Dec 2 at 17:10










            2




            2




            THIS is the rigorous way (+1)
            – Robert Z
            Dec 2 at 17:10






            THIS is the rigorous way (+1)
            – Robert Z
            Dec 2 at 17:10













            3














            Note that



            $$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$






            share|cite|improve this answer





















            • The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
              – Clement C.
              Dec 1 at 18:35


















            3














            Note that



            $$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$






            share|cite|improve this answer





















            • The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
              – Clement C.
              Dec 1 at 18:35
















            3












            3








            3






            Note that



            $$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$






            share|cite|improve this answer












            Note that



            $$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 18:32









            KM101

            4,528418




            4,528418












            • The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
              – Clement C.
              Dec 1 at 18:35




















            • The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
              – Clement C.
              Dec 1 at 18:35


















            The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
            – Clement C.
            Dec 1 at 18:35






            The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
            – Clement C.
            Dec 1 at 18:35













            0














            Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
            $$
            g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
            $$
            since
            $$
            frac{g(n)}{f(n)}to0
            $$
            as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
            $$
            lim_{xto0}frac{log (1+x)-0}{x-0}=1
            $$
            by definition of the derivative.






            share|cite|improve this answer





















            • So the result is 0*1 ?
              – Samy Bencherif
              Dec 2 at 6:42










            • No to the limit is $exp(0times 1)$
              – Foobaz John
              Dec 2 at 15:57


















            0














            Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
            $$
            g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
            $$
            since
            $$
            frac{g(n)}{f(n)}to0
            $$
            as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
            $$
            lim_{xto0}frac{log (1+x)-0}{x-0}=1
            $$
            by definition of the derivative.






            share|cite|improve this answer





















            • So the result is 0*1 ?
              – Samy Bencherif
              Dec 2 at 6:42










            • No to the limit is $exp(0times 1)$
              – Foobaz John
              Dec 2 at 15:57
















            0












            0








            0






            Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
            $$
            g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
            $$
            since
            $$
            frac{g(n)}{f(n)}to0
            $$
            as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
            $$
            lim_{xto0}frac{log (1+x)-0}{x-0}=1
            $$
            by definition of the derivative.






            share|cite|improve this answer












            Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
            $$
            g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
            $$
            since
            $$
            frac{g(n)}{f(n)}to0
            $$
            as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
            $$
            lim_{xto0}frac{log (1+x)-0}{x-0}=1
            $$
            by definition of the derivative.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 18:38









            Foobaz John

            20.9k41250




            20.9k41250












            • So the result is 0*1 ?
              – Samy Bencherif
              Dec 2 at 6:42










            • No to the limit is $exp(0times 1)$
              – Foobaz John
              Dec 2 at 15:57




















            • So the result is 0*1 ?
              – Samy Bencherif
              Dec 2 at 6:42










            • No to the limit is $exp(0times 1)$
              – Foobaz John
              Dec 2 at 15:57


















            So the result is 0*1 ?
            – Samy Bencherif
            Dec 2 at 6:42




            So the result is 0*1 ?
            – Samy Bencherif
            Dec 2 at 6:42












            No to the limit is $exp(0times 1)$
            – Foobaz John
            Dec 2 at 15:57






            No to the limit is $exp(0times 1)$
            – Foobaz John
            Dec 2 at 15:57




















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