Could someone tell me why $log n 0$? [on hold]Can someone explain log?How can I calculate $lim_x to 0 log(cos(x))/log(cos(3x))$ without l'Hopital?Proving limit with $log(n!)$Simplyfing Log QuestionHow to compute $=lim_n to infty Big( fraclog(n+1)log(n) cdot fracn-2n-1 Big)$“by hand”?Why does $(log n)^log n = Omega(n^10)$?Why do $2log x$ and $log x^2$ look different when graphing?Value of limit $(1+e^-x)^2^x logx$Find $lim_nrightarrowinfty fraclog(2 + 3^n)2n$Solving $2log(n!) - nlog n le 2nlog n$

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Could someone tell me why $log n 0$? [on hold]


Can someone explain log?How can I calculate $lim_x to 0 log(cos(x))/log(cos(3x))$ without l'Hopital?Proving limit with $log(n!)$Simplyfing Log QuestionHow to compute $=lim_n to infty Big( fraclog(n+1)log(n) cdot fracn-2n-1 Big)$“by hand”?Why does $(log n)^log n = Omega(n^10)$?Why do $2log x$ and $log x^2$ look different when graphing?Value of limit $(1+e^-x)^2^x logx$Find $lim_nrightarrowinfty fraclog(2 + 3^n)2n$Solving $2log(n!) - nlog n le 2nlog n$













-3












$begingroup$



To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
$$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$




Could some one tell me how to convert the expression 1 to 2?
I don't understand.
Thank you!










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WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03


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." – RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.




















    -3












    $begingroup$



    To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
    $$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$




    Could some one tell me how to convert the expression 1 to 2?
    I don't understand.
    Thank you!










    share|cite|improve this question









    New contributor




    WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$



    put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03


    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." – RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      -3












      -3








      -3





      $begingroup$



      To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
      $$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$




      Could some one tell me how to convert the expression 1 to 2?
      I don't understand.
      Thank you!










      share|cite|improve this question









      New contributor




      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
      $$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$




      Could some one tell me how to convert the expression 1 to 2?
      I don't understand.
      Thank you!







      logarithms






      share|cite|improve this question









      New contributor




      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited Mar 11 at 13:34









      Rócherz

      2,9512821




      2,9512821






      New contributor




      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked Mar 11 at 10:18









      WU NICKWU NICK

      1




      1




      New contributor




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





      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03


      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." – RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會
      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03


      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." – RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
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          active

          oldest

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          $begingroup$

          If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$






          share|cite|improve this answer









          $endgroup$



















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$






                share|cite|improve this answer









                $endgroup$



                If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 11 at 10:25









                FredFred

                48.3k1849




                48.3k1849













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