Find the limit of $e^x/2^x$ as $x$ approaches infinity












2












$begingroup$


I am trying to find the asymptotic relation between $e^x$ and $2^x$. I tried to use limit comparison: $$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$



I tried to use L'Hopital's rule:



$$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$
$$=lim_{xtoinfty}left(frac{e^x}{ln(2) cdot 2^x}right)$$



which doesn't really help. Is there any way to compute this limit? Thanks!






limits







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$endgroup$








  • 3




    $begingroup$
    Try using $a^x=e^{xln(a)}$
    $endgroup$
    – nathan.j.mcdougall
    Jan 18 at 3:22








  • 3




    $begingroup$
    $frac{e}{2}gt 1$
    $endgroup$
    – John Douma
    Jan 18 at 3:28






  • 2




    $begingroup$
    Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
    $endgroup$
    – MJD
    Jan 18 at 3:32


















2












$begingroup$


I am trying to find the asymptotic relation between $e^x$ and $2^x$. I tried to use limit comparison: $$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$



I tried to use L'Hopital's rule:



$$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$
$$=lim_{xtoinfty}left(frac{e^x}{ln(2) cdot 2^x}right)$$



which doesn't really help. Is there any way to compute this limit? Thanks!






limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Try using $a^x=e^{xln(a)}$
    $endgroup$
    – nathan.j.mcdougall
    Jan 18 at 3:22








  • 3




    $begingroup$
    $frac{e}{2}gt 1$
    $endgroup$
    – John Douma
    Jan 18 at 3:28






  • 2




    $begingroup$
    Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
    $endgroup$
    – MJD
    Jan 18 at 3:32
















2












2








2





$begingroup$


I am trying to find the asymptotic relation between $e^x$ and $2^x$. I tried to use limit comparison: $$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$



I tried to use L'Hopital's rule:



$$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$
$$=lim_{xtoinfty}left(frac{e^x}{ln(2) cdot 2^x}right)$$



which doesn't really help. Is there any way to compute this limit? Thanks!






limits







share|cite|improve this question











$endgroup$




I am trying to find the asymptotic relation between $e^x$ and $2^x$. I tried to use limit comparison: $$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$



I tried to use L'Hopital's rule:



$$lim_{xtoinfty}left(frac{e^x}{2^x}right)$$
$$=lim_{xtoinfty}left(frac{e^x}{ln(2) cdot 2^x}right)$$



which doesn't really help. Is there any way to compute this limit? Thanks!






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 21:08









Antonio Vargas

20.7k245111




20.7k245111










asked Jan 18 at 3:20









Yifei HeYifei He

1286




1286








  • 3




    $begingroup$
    Try using $a^x=e^{xln(a)}$
    $endgroup$
    – nathan.j.mcdougall
    Jan 18 at 3:22








  • 3




    $begingroup$
    $frac{e}{2}gt 1$
    $endgroup$
    – John Douma
    Jan 18 at 3:28






  • 2




    $begingroup$
    Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
    $endgroup$
    – MJD
    Jan 18 at 3:32
















  • 3




    $begingroup$
    Try using $a^x=e^{xln(a)}$
    $endgroup$
    – nathan.j.mcdougall
    Jan 18 at 3:22








  • 3




    $begingroup$
    $frac{e}{2}gt 1$
    $endgroup$
    – John Douma
    Jan 18 at 3:28






  • 2




    $begingroup$
    Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
    $endgroup$
    – MJD
    Jan 18 at 3:32










3




3




$begingroup$
Try using $a^x=e^{xln(a)}$
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:22






$begingroup$
Try using $a^x=e^{xln(a)}$
$endgroup$
– nathan.j.mcdougall
Jan 18 at 3:22






3




3




$begingroup$
$frac{e}{2}gt 1$
$endgroup$
– John Douma
Jan 18 at 3:28




$begingroup$
$frac{e}{2}gt 1$
$endgroup$
– John Douma
Jan 18 at 3:28




2




2




$begingroup$
Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
$endgroup$
– MJD
Jan 18 at 3:32






$begingroup$
Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=frac1{ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so...
$endgroup$
– MJD
Jan 18 at 3:32












2 Answers
2






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5












$begingroup$

Note that



$$lim_{x to infty} left(cfrac{e^x}{2^x}right) = lim_{x to infty} left(cfrac{e}{2}right)^x = infty tag{1}label{eq1}$$



because $e > 2$ so $frac{e}{2} > 1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yeah, it can't get any simpler than this approach.
    $endgroup$
    – Randall
    Jan 18 at 3:29



















3












$begingroup$

HINT



$$frac{e^x}{2^x}=frac{e^x}{e^{xln2}}=e^{x(1-ln 2)}$$



Then:



$$lim_{xtoinfty}{x(1-ln 2)}= ?$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    5












    $begingroup$

    Note that



    $$lim_{x to infty} left(cfrac{e^x}{2^x}right) = lim_{x to infty} left(cfrac{e}{2}right)^x = infty tag{1}label{eq1}$$



    because $e > 2$ so $frac{e}{2} > 1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Yeah, it can't get any simpler than this approach.
      $endgroup$
      – Randall
      Jan 18 at 3:29
















    5












    $begingroup$

    Note that



    $$lim_{x to infty} left(cfrac{e^x}{2^x}right) = lim_{x to infty} left(cfrac{e}{2}right)^x = infty tag{1}label{eq1}$$



    because $e > 2$ so $frac{e}{2} > 1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Yeah, it can't get any simpler than this approach.
      $endgroup$
      – Randall
      Jan 18 at 3:29














    5












    5








    5





    $begingroup$

    Note that



    $$lim_{x to infty} left(cfrac{e^x}{2^x}right) = lim_{x to infty} left(cfrac{e}{2}right)^x = infty tag{1}label{eq1}$$



    because $e > 2$ so $frac{e}{2} > 1$.






    share|cite|improve this answer









    $endgroup$



    Note that



    $$lim_{x to infty} left(cfrac{e^x}{2^x}right) = lim_{x to infty} left(cfrac{e}{2}right)^x = infty tag{1}label{eq1}$$



    because $e > 2$ so $frac{e}{2} > 1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 at 3:28









    John OmielanJohn Omielan

    2,626212




    2,626212








    • 1




      $begingroup$
      Yeah, it can't get any simpler than this approach.
      $endgroup$
      – Randall
      Jan 18 at 3:29














    • 1




      $begingroup$
      Yeah, it can't get any simpler than this approach.
      $endgroup$
      – Randall
      Jan 18 at 3:29








    1




    1




    $begingroup$
    Yeah, it can't get any simpler than this approach.
    $endgroup$
    – Randall
    Jan 18 at 3:29




    $begingroup$
    Yeah, it can't get any simpler than this approach.
    $endgroup$
    – Randall
    Jan 18 at 3:29











    3












    $begingroup$

    HINT



    $$frac{e^x}{2^x}=frac{e^x}{e^{xln2}}=e^{x(1-ln 2)}$$



    Then:



    $$lim_{xtoinfty}{x(1-ln 2)}= ?$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      HINT



      $$frac{e^x}{2^x}=frac{e^x}{e^{xln2}}=e^{x(1-ln 2)}$$



      Then:



      $$lim_{xtoinfty}{x(1-ln 2)}= ?$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        HINT



        $$frac{e^x}{2^x}=frac{e^x}{e^{xln2}}=e^{x(1-ln 2)}$$



        Then:



        $$lim_{xtoinfty}{x(1-ln 2)}= ?$$






        share|cite|improve this answer









        $endgroup$



        HINT



        $$frac{e^x}{2^x}=frac{e^x}{e^{xln2}}=e^{x(1-ln 2)}$$



        Then:



        $$lim_{xtoinfty}{x(1-ln 2)}= ?$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 3:26









        Rhys HughesRhys Hughes

        6,4041530




        6,4041530






























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