How do I calculate $lim_{xto infty} sqrt{(2pi x)}^frac{1}{x}$












3












$begingroup$


I just started studying limits a week ago and today I got this question on my YouTube feed. I am having a hard time with it.



The question is:




Find the value of: $lim_{xto infty}(frac{x!}{x^x})^frac{1}{x}$




The answer to this question is $frac{1}{e}$



There in the comment section, someone suggested to use the Stirling's approximation. It states




$n! approx e^{-n}*n^n*sqrt{2pi n}$




After putting this in the question it reduces down to: proving that $lim_{xto infty} sqrt{(2pi x)}^frac{1}{x}=1$



I just don’t know how to eliminate that $pi$ in the expression. Wolfram Alpha suggested to use Puiseux series. Apparently, the wiki page is just too advanced for me to comprehend.



I would be really thankful to anyone who could shed some light on this problem!



( The video link is: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=0s )






calculus limits







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I just started studying limits a week ago and today I got this question on my YouTube feed. I am having a hard time with it.



    The question is:




    Find the value of: $lim_{xto infty}(frac{x!}{x^x})^frac{1}{x}$




    The answer to this question is $frac{1}{e}$



    There in the comment section, someone suggested to use the Stirling's approximation. It states




    $n! approx e^{-n}*n^n*sqrt{2pi n}$




    After putting this in the question it reduces down to: proving that $lim_{xto infty} sqrt{(2pi x)}^frac{1}{x}=1$



    I just don’t know how to eliminate that $pi$ in the expression. Wolfram Alpha suggested to use Puiseux series. Apparently, the wiki page is just too advanced for me to comprehend.



    I would be really thankful to anyone who could shed some light on this problem!



    ( The video link is: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=0s )






    calculus limits







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I just started studying limits a week ago and today I got this question on my YouTube feed. I am having a hard time with it.



      The question is:




      Find the value of: $lim_{xto infty}(frac{x!}{x^x})^frac{1}{x}$




      The answer to this question is $frac{1}{e}$



      There in the comment section, someone suggested to use the Stirling's approximation. It states




      $n! approx e^{-n}*n^n*sqrt{2pi n}$




      After putting this in the question it reduces down to: proving that $lim_{xto infty} sqrt{(2pi x)}^frac{1}{x}=1$



      I just don’t know how to eliminate that $pi$ in the expression. Wolfram Alpha suggested to use Puiseux series. Apparently, the wiki page is just too advanced for me to comprehend.



      I would be really thankful to anyone who could shed some light on this problem!



      ( The video link is: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=0s )






      calculus limits







      share|cite|improve this question









      $endgroup$




      I just started studying limits a week ago and today I got this question on my YouTube feed. I am having a hard time with it.



      The question is:




      Find the value of: $lim_{xto infty}(frac{x!}{x^x})^frac{1}{x}$




      The answer to this question is $frac{1}{e}$



      There in the comment section, someone suggested to use the Stirling's approximation. It states




      $n! approx e^{-n}*n^n*sqrt{2pi n}$




      After putting this in the question it reduces down to: proving that $lim_{xto infty} sqrt{(2pi x)}^frac{1}{x}=1$



      I just don’t know how to eliminate that $pi$ in the expression. Wolfram Alpha suggested to use Puiseux series. Apparently, the wiki page is just too advanced for me to comprehend.



      I would be really thankful to anyone who could shed some light on this problem!



      ( The video link is: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=0s )






      calculus limits




      calculus limits






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 22 '18 at 12:10









      Aaryan DewanAaryan Dewan

      37639




      37639






















          1 Answer
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          5












          $begingroup$

          Why on earth would you need to eliminate $pi$? This is actually a relatively simple limit. The easiest way I see to do this is to rewrite it as
          $$lim_{xtoinfty} (2pi x)^{frac 1{2x}}$$
          We can then compute
          $$lim_{xtoinfty} log left((2pi x)^{frac 1{2x}}right)=lim_{xtoinfty} frac{1}{2x}log 2pi x$$
          If you truly want, you can split this as
          $$lim_{xtoinfty} frac{1}{2x}(log 2pi+log x)$$
          The limit is $0$, hence the original limit is $1$. Plugging it into the approximation gives you $e^{-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
            $endgroup$
            – Kemono Chen
            Dec 22 '18 at 12:31












          • $begingroup$
            @Kemono Sure, probably. Changed it. Thanks.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:32










          • $begingroup$
            Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
            $endgroup$
            – Aaryan Dewan
            Dec 22 '18 at 12:39










          • $begingroup$
            @Aaryan You're welcome.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:40











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Why on earth would you need to eliminate $pi$? This is actually a relatively simple limit. The easiest way I see to do this is to rewrite it as
          $$lim_{xtoinfty} (2pi x)^{frac 1{2x}}$$
          We can then compute
          $$lim_{xtoinfty} log left((2pi x)^{frac 1{2x}}right)=lim_{xtoinfty} frac{1}{2x}log 2pi x$$
          If you truly want, you can split this as
          $$lim_{xtoinfty} frac{1}{2x}(log 2pi+log x)$$
          The limit is $0$, hence the original limit is $1$. Plugging it into the approximation gives you $e^{-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
            $endgroup$
            – Kemono Chen
            Dec 22 '18 at 12:31












          • $begingroup$
            @Kemono Sure, probably. Changed it. Thanks.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:32










          • $begingroup$
            Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
            $endgroup$
            – Aaryan Dewan
            Dec 22 '18 at 12:39










          • $begingroup$
            @Aaryan You're welcome.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:40
















          5












          $begingroup$

          Why on earth would you need to eliminate $pi$? This is actually a relatively simple limit. The easiest way I see to do this is to rewrite it as
          $$lim_{xtoinfty} (2pi x)^{frac 1{2x}}$$
          We can then compute
          $$lim_{xtoinfty} log left((2pi x)^{frac 1{2x}}right)=lim_{xtoinfty} frac{1}{2x}log 2pi x$$
          If you truly want, you can split this as
          $$lim_{xtoinfty} frac{1}{2x}(log 2pi+log x)$$
          The limit is $0$, hence the original limit is $1$. Plugging it into the approximation gives you $e^{-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
            $endgroup$
            – Kemono Chen
            Dec 22 '18 at 12:31












          • $begingroup$
            @Kemono Sure, probably. Changed it. Thanks.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:32










          • $begingroup$
            Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
            $endgroup$
            – Aaryan Dewan
            Dec 22 '18 at 12:39










          • $begingroup$
            @Aaryan You're welcome.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:40














          5












          5








          5





          $begingroup$

          Why on earth would you need to eliminate $pi$? This is actually a relatively simple limit. The easiest way I see to do this is to rewrite it as
          $$lim_{xtoinfty} (2pi x)^{frac 1{2x}}$$
          We can then compute
          $$lim_{xtoinfty} log left((2pi x)^{frac 1{2x}}right)=lim_{xtoinfty} frac{1}{2x}log 2pi x$$
          If you truly want, you can split this as
          $$lim_{xtoinfty} frac{1}{2x}(log 2pi+log x)$$
          The limit is $0$, hence the original limit is $1$. Plugging it into the approximation gives you $e^{-1}$.






          share|cite|improve this answer











          $endgroup$



          Why on earth would you need to eliminate $pi$? This is actually a relatively simple limit. The easiest way I see to do this is to rewrite it as
          $$lim_{xtoinfty} (2pi x)^{frac 1{2x}}$$
          We can then compute
          $$lim_{xtoinfty} log left((2pi x)^{frac 1{2x}}right)=lim_{xtoinfty} frac{1}{2x}log 2pi x$$
          If you truly want, you can split this as
          $$lim_{xtoinfty} frac{1}{2x}(log 2pi+log x)$$
          The limit is $0$, hence the original limit is $1$. Plugging it into the approximation gives you $e^{-1}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 22 '18 at 12:32

























          answered Dec 22 '18 at 12:18









          Matt SamuelMatt Samuel

          38.7k63769




          38.7k63769












          • $begingroup$
            In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
            $endgroup$
            – Kemono Chen
            Dec 22 '18 at 12:31












          • $begingroup$
            @Kemono Sure, probably. Changed it. Thanks.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:32










          • $begingroup$
            Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
            $endgroup$
            – Aaryan Dewan
            Dec 22 '18 at 12:39










          • $begingroup$
            @Aaryan You're welcome.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:40


















          • $begingroup$
            In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
            $endgroup$
            – Kemono Chen
            Dec 22 '18 at 12:31












          • $begingroup$
            @Kemono Sure, probably. Changed it. Thanks.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:32










          • $begingroup$
            Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
            $endgroup$
            – Aaryan Dewan
            Dec 22 '18 at 12:39










          • $begingroup$
            @Aaryan You're welcome.
            $endgroup$
            – Matt Samuel
            Dec 22 '18 at 12:40
















          $begingroup$
          In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
          $endgroup$
          – Kemono Chen
          Dec 22 '18 at 12:31






          $begingroup$
          In order to avoid ambiguous, I believe using $logleft((2pi x)^{frac1{2x}}right)$ is better
          $endgroup$
          – Kemono Chen
          Dec 22 '18 at 12:31














          $begingroup$
          @Kemono Sure, probably. Changed it. Thanks.
          $endgroup$
          – Matt Samuel
          Dec 22 '18 at 12:32




          $begingroup$
          @Kemono Sure, probably. Changed it. Thanks.
          $endgroup$
          – Matt Samuel
          Dec 22 '18 at 12:32












          $begingroup$
          Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
          $endgroup$
          – Aaryan Dewan
          Dec 22 '18 at 12:39




          $begingroup$
          Holy Moly! I forgot that $lim_{xto infty}(frac{log(x)}{x})=0$ ! . Thank You so much :)
          $endgroup$
          – Aaryan Dewan
          Dec 22 '18 at 12:39












          $begingroup$
          @Aaryan You're welcome.
          $endgroup$
          – Matt Samuel
          Dec 22 '18 at 12:40




          $begingroup$
          @Aaryan You're welcome.
          $endgroup$
          – Matt Samuel
          Dec 22 '18 at 12:40


















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