find $lim_xrightarrow infty fraca^xx^alpha$ [duplicate]

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  • How to prove that exponential grows faster than polynomial?

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Given that $a> 1$, $alpha in mathbbR $,then find
$$lim_xrightarrow infty fraca^xx^alpha$$



My attempts :



$fraca^xx^alpha le frac1x^alpha$



now i take $lim_xrightarrow infty frac1x^alpha = frac1infty = 0$



Is its correct ??



thanks in advance




real-analysis




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    Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
    – Mattos
    Aug 13 at 7:39











  • let me thinks more
    – stupid
    Aug 13 at 7:50










  • Related math.stackexchange.com/questions/2489665/…
    – rtybase
    Aug 13 at 7:55














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This question already has an answer here:



  • How to prove that exponential grows faster than polynomial?

    13 answers



Given that $a> 1$, $alpha in mathbbR $,then find
$$lim_xrightarrow infty fraca^xx^alpha$$



My attempts :



$fraca^xx^alpha le frac1x^alpha$



now i take $lim_xrightarrow infty frac1x^alpha = frac1infty = 0$



Is its correct ??



thanks in advance




real-analysis




share|cite|improve this question












marked as duplicate by Hans Lundmark, RRL real-analysis
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 Aug 13 at 13:12


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  • 1




    Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
    – Mattos
    Aug 13 at 7:39











  • let me thinks more
    – stupid
    Aug 13 at 7:50










  • Related math.stackexchange.com/questions/2489665/…
    – rtybase
    Aug 13 at 7:55












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite












This question already has an answer here:



  • How to prove that exponential grows faster than polynomial?

    13 answers



Given that $a> 1$, $alpha in mathbbR $,then find
$$lim_xrightarrow infty fraca^xx^alpha$$



My attempts :



$fraca^xx^alpha le frac1x^alpha$



now i take $lim_xrightarrow infty frac1x^alpha = frac1infty = 0$



Is its correct ??



thanks in advance




real-analysis




share|cite|improve this question













This question already has an answer here:



  • How to prove that exponential grows faster than polynomial?

    13 answers



Given that $a> 1$, $alpha in mathbbR $,then find
$$lim_xrightarrow infty fraca^xx^alpha$$



My attempts :



$fraca^xx^alpha le frac1x^alpha$



now i take $lim_xrightarrow infty frac1x^alpha = frac1infty = 0$



Is its correct ??



thanks in advance





This question already has an answer here:



  • How to prove that exponential grows faster than polynomial?

    13 answers





real-analysis





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 13 at 7:34









stupid

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marked as duplicate by Hans Lundmark, RRL real-analysis
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 Aug 13 at 13:12


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
    – Mattos
    Aug 13 at 7:39











  • let me thinks more
    – stupid
    Aug 13 at 7:50










  • Related math.stackexchange.com/questions/2489665/…
    – rtybase
    Aug 13 at 7:55












  • 1




    Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
    – Mattos
    Aug 13 at 7:39











  • let me thinks more
    – stupid
    Aug 13 at 7:50










  • Related math.stackexchange.com/questions/2489665/…
    – rtybase
    Aug 13 at 7:55







1




1




Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
– Mattos
Aug 13 at 7:39





Why is $a^x/x^alpha le 1/x^alpha implies a^x le 1$? $a^x$ is increasing for $a > 1$.
– Mattos
Aug 13 at 7:39













let me thinks more
– stupid
Aug 13 at 7:50




let me thinks more
– stupid
Aug 13 at 7:50












Related math.stackexchange.com/questions/2489665/…
– rtybase
Aug 13 at 7:55




Related math.stackexchange.com/questions/2489665/…
– rtybase
Aug 13 at 7:55










1 Answer
1






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up vote
2
down vote



accepted










As pointed out by Mattos your first inequality is wrong. $frac a^x x^alpha=frac e^xlog a x^alpha geq frac (x log a)^n (n!)x^alpha to infty$ if we take $n > alpha $.






share|cite|improve this answer



























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    As pointed out by Mattos your first inequality is wrong. $frac a^x x^alpha=frac e^xlog a x^alpha geq frac (x log a)^n (n!)x^alpha to infty$ if we take $n > alpha $.






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      As pointed out by Mattos your first inequality is wrong. $frac a^x x^alpha=frac e^xlog a x^alpha geq frac (x log a)^n (n!)x^alpha to infty$ if we take $n > alpha $.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        As pointed out by Mattos your first inequality is wrong. $frac a^x x^alpha=frac e^xlog a x^alpha geq frac (x log a)^n (n!)x^alpha to infty$ if we take $n > alpha $.






        share|cite|improve this answer












        As pointed out by Mattos your first inequality is wrong. $frac a^x x^alpha=frac e^xlog a x^alpha geq frac (x log a)^n (n!)x^alpha to infty$ if we take $n > alpha $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 13 at 7:51









        Kavi Rama Murthy

        22.2k2933




        22.2k2933












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