Show that $lim_{ntoinfty}frac{log_an}{n} = 0$ for $0<a<1$












2












$begingroup$



Let $0<a<1$, prove that:
$$
lim_{ntoinfty}frac{log_an}{n} = 0
$$




I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
$$
frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
$$



We know that for $a>1$ and $kin mathbb N$:
$$
lim_{ntoinfty}frac{n^k}{a^n}=0
$$
Using that fact we may show that:
$$
exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
$$



Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:



$$
lim_{ntoinfty} frac{log_an}{n} = 0
$$



Now consider the case for $0<a<1$:
$$
left|frac{log_an}{n}right| < varepsilon
$$



And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?



Also one of my thoughts was to present $a$ as:
$$
a = frac{1}{1+r} , r in mathbb R
$$



and then try to use Bernoulli's, but that didn't yield anything I could use.



Please note that i'm free to use anything before the definition of a derivative.



Update



Based on the hint by MathLover I think this is how I can proceed. Use the fact that:



$$
log_ax = - log_{1/a}x
$$



Hence:
$$
log_{1/a}n < nvarepsilon
$$



Define $b = {1 over a} > 1$, so:



$$
n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
$$



Now based on the reasoning for case when $a > 1$ we may as well conclude that:
$$
exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
$$



Thus:
$$
begin{cases}
lim_{ntoinfty}frac{log_an}{n} = 0 \
ain (0, 1) bigcup (1, +infty)
end{cases}
$$






calculus limits logarithms limits-without-lhopital epsilon-delta







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $0<a<1$, prove that:
    $$
    lim_{ntoinfty}frac{log_an}{n} = 0
    $$




    I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
    $$
    frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
    $$



    We know that for $a>1$ and $kin mathbb N$:
    $$
    lim_{ntoinfty}frac{n^k}{a^n}=0
    $$
    Using that fact we may show that:
    $$
    exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
    $$



    Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:



    $$
    lim_{ntoinfty} frac{log_an}{n} = 0
    $$



    Now consider the case for $0<a<1$:
    $$
    left|frac{log_an}{n}right| < varepsilon
    $$



    And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?



    Also one of my thoughts was to present $a$ as:
    $$
    a = frac{1}{1+r} , r in mathbb R
    $$



    and then try to use Bernoulli's, but that didn't yield anything I could use.



    Please note that i'm free to use anything before the definition of a derivative.



    Update



    Based on the hint by MathLover I think this is how I can proceed. Use the fact that:



    $$
    log_ax = - log_{1/a}x
    $$



    Hence:
    $$
    log_{1/a}n < nvarepsilon
    $$



    Define $b = {1 over a} > 1$, so:



    $$
    n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
    $$



    Now based on the reasoning for case when $a > 1$ we may as well conclude that:
    $$
    exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
    $$



    Thus:
    $$
    begin{cases}
    lim_{ntoinfty}frac{log_an}{n} = 0 \
    ain (0, 1) bigcup (1, +infty)
    end{cases}
    $$






    calculus limits logarithms limits-without-lhopital epsilon-delta







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $0<a<1$, prove that:
      $$
      lim_{ntoinfty}frac{log_an}{n} = 0
      $$




      I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
      $$
      frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
      $$



      We know that for $a>1$ and $kin mathbb N$:
      $$
      lim_{ntoinfty}frac{n^k}{a^n}=0
      $$
      Using that fact we may show that:
      $$
      exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
      $$



      Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:



      $$
      lim_{ntoinfty} frac{log_an}{n} = 0
      $$



      Now consider the case for $0<a<1$:
      $$
      left|frac{log_an}{n}right| < varepsilon
      $$



      And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?



      Also one of my thoughts was to present $a$ as:
      $$
      a = frac{1}{1+r} , r in mathbb R
      $$



      and then try to use Bernoulli's, but that didn't yield anything I could use.



      Please note that i'm free to use anything before the definition of a derivative.



      Update



      Based on the hint by MathLover I think this is how I can proceed. Use the fact that:



      $$
      log_ax = - log_{1/a}x
      $$



      Hence:
      $$
      log_{1/a}n < nvarepsilon
      $$



      Define $b = {1 over a} > 1$, so:



      $$
      n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
      $$



      Now based on the reasoning for case when $a > 1$ we may as well conclude that:
      $$
      exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
      $$



      Thus:
      $$
      begin{cases}
      lim_{ntoinfty}frac{log_an}{n} = 0 \
      ain (0, 1) bigcup (1, +infty)
      end{cases}
      $$






      calculus limits logarithms limits-without-lhopital epsilon-delta







      share|cite|improve this question











      $endgroup$





      Let $0<a<1$, prove that:
      $$
      lim_{ntoinfty}frac{log_an}{n} = 0
      $$




      I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
      $$
      frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
      $$



      We know that for $a>1$ and $kin mathbb N$:
      $$
      lim_{ntoinfty}frac{n^k}{a^n}=0
      $$
      Using that fact we may show that:
      $$
      exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
      $$



      Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:



      $$
      lim_{ntoinfty} frac{log_an}{n} = 0
      $$



      Now consider the case for $0<a<1$:
      $$
      left|frac{log_an}{n}right| < varepsilon
      $$



      And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?



      Also one of my thoughts was to present $a$ as:
      $$
      a = frac{1}{1+r} , r in mathbb R
      $$



      and then try to use Bernoulli's, but that didn't yield anything I could use.



      Please note that i'm free to use anything before the definition of a derivative.



      Update



      Based on the hint by MathLover I think this is how I can proceed. Use the fact that:



      $$
      log_ax = - log_{1/a}x
      $$



      Hence:
      $$
      log_{1/a}n < nvarepsilon
      $$



      Define $b = {1 over a} > 1$, so:



      $$
      n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
      $$



      Now based on the reasoning for case when $a > 1$ we may as well conclude that:
      $$
      exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
      $$



      Thus:
      $$
      begin{cases}
      lim_{ntoinfty}frac{log_an}{n} = 0 \
      ain (0, 1) bigcup (1, +infty)
      end{cases}
      $$






      calculus limits logarithms limits-without-lhopital epsilon-delta




      calculus limits logarithms limits-without-lhopital epsilon-delta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 17:39







      roman

















      asked Dec 4 '18 at 17:07









      romanroman

      2,17321224




      2,17321224






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$



          So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$



          (If you aren't convinced of the last step, see this)



          As such, if the limit exists, it has the value of 0.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint:
            $$log_a(n) = -log_{1/a}(n).$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$



                So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$



                (If you aren't convinced of the last step, see this)



                As such, if the limit exists, it has the value of 0.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$



                  So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$



                  (If you aren't convinced of the last step, see this)



                  As such, if the limit exists, it has the value of 0.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$



                    So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$



                    (If you aren't convinced of the last step, see this)



                    As such, if the limit exists, it has the value of 0.






                    share|cite|improve this answer









                    $endgroup$



                    We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$



                    So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$



                    (If you aren't convinced of the last step, see this)



                    As such, if the limit exists, it has the value of 0.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 '18 at 17:16









                    Don ThousandDon Thousand

                    4,332734




                    4,332734























                        1












                        $begingroup$

                        Hint:
                        $$log_a(n) = -log_{1/a}(n).$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint:
                          $$log_a(n) = -log_{1/a}(n).$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:
                            $$log_a(n) = -log_{1/a}(n).$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:
                            $$log_a(n) = -log_{1/a}(n).$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 4 '18 at 17:18









                            Math LoverMath Lover

                            14k31436




                            14k31436























                                0












                                $begingroup$

                                Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 4 '18 at 17:43









                                    Paramanand SinghParamanand Singh

                                    49.9k556163




                                    49.9k556163






























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