Computing limit of $f(x) = x^{(frac{1}{x} - 1)}$











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Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .



My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .






calculus real-analysis limits limits-without-lhopital







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  • The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
    – Paramanand Singh
    Nov 20 at 2:01










  • @ParamanandSingh How we can prove that statement ?
    – S.H.W
    Nov 20 at 3:16










  • Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
    – Paramanand Singh
    Nov 20 at 5:03












  • @ParamanandSingh Thanks a lot .
    – S.H.W
    Nov 20 at 17:49















up vote
0
down vote

favorite












Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .



My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .






calculus real-analysis limits limits-without-lhopital







share|cite|improve this question






















  • The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
    – Paramanand Singh
    Nov 20 at 2:01










  • @ParamanandSingh How we can prove that statement ?
    – S.H.W
    Nov 20 at 3:16










  • Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
    – Paramanand Singh
    Nov 20 at 5:03












  • @ParamanandSingh Thanks a lot .
    – S.H.W
    Nov 20 at 17:49













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .



My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .






calculus real-analysis limits limits-without-lhopital







share|cite|improve this question













Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .



My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .






calculus real-analysis limits limits-without-lhopital




calculus real-analysis limits limits-without-lhopital






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 0:42









S.H.W

1,1491922




1,1491922












  • The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
    – Paramanand Singh
    Nov 20 at 2:01










  • @ParamanandSingh How we can prove that statement ?
    – S.H.W
    Nov 20 at 3:16










  • Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
    – Paramanand Singh
    Nov 20 at 5:03












  • @ParamanandSingh Thanks a lot .
    – S.H.W
    Nov 20 at 17:49


















  • The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
    – Paramanand Singh
    Nov 20 at 2:01










  • @ParamanandSingh How we can prove that statement ?
    – S.H.W
    Nov 20 at 3:16










  • Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
    – Paramanand Singh
    Nov 20 at 5:03












  • @ParamanandSingh Thanks a lot .
    – S.H.W
    Nov 20 at 17:49
















The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 at 2:01




The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 at 2:01












@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 at 3:16




@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 at 3:16












Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 at 5:03






Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 at 5:03














@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 at 17:49




@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 at 17:49










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










This is not indeterminate when $xto 0^+$:





  • $ln xto -infty$,


  • $1-xto 1$, so $(1-x)ln xto-infty$

  • the denominator tends to $0^+$
    so the exponent tends to $-infty$, and the exponential indeed tends to $0$.






share|cite|improve this answer





















  • How we can make it in a rigorous way ?
    – S.H.W
    Nov 20 at 0:52


















up vote
1
down vote













$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$



$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$



$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$



$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$






share|cite|improve this answer




























    up vote
    1
    down vote













    Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      This is not indeterminate when $xto 0^+$:





      • $ln xto -infty$,


      • $1-xto 1$, so $(1-x)ln xto-infty$

      • the denominator tends to $0^+$
        so the exponent tends to $-infty$, and the exponential indeed tends to $0$.






      share|cite|improve this answer





















      • How we can make it in a rigorous way ?
        – S.H.W
        Nov 20 at 0:52















      up vote
      2
      down vote



      accepted










      This is not indeterminate when $xto 0^+$:





      • $ln xto -infty$,


      • $1-xto 1$, so $(1-x)ln xto-infty$

      • the denominator tends to $0^+$
        so the exponent tends to $-infty$, and the exponential indeed tends to $0$.






      share|cite|improve this answer





















      • How we can make it in a rigorous way ?
        – S.H.W
        Nov 20 at 0:52













      up vote
      2
      down vote



      accepted







      up vote
      2
      down vote



      accepted






      This is not indeterminate when $xto 0^+$:





      • $ln xto -infty$,


      • $1-xto 1$, so $(1-x)ln xto-infty$

      • the denominator tends to $0^+$
        so the exponent tends to $-infty$, and the exponential indeed tends to $0$.






      share|cite|improve this answer












      This is not indeterminate when $xto 0^+$:





      • $ln xto -infty$,


      • $1-xto 1$, so $(1-x)ln xto-infty$

      • the denominator tends to $0^+$
        so the exponent tends to $-infty$, and the exponential indeed tends to $0$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 20 at 0:51









      Bernard

      117k637109




      117k637109












      • How we can make it in a rigorous way ?
        – S.H.W
        Nov 20 at 0:52


















      • How we can make it in a rigorous way ?
        – S.H.W
        Nov 20 at 0:52
















      How we can make it in a rigorous way ?
      – S.H.W
      Nov 20 at 0:52




      How we can make it in a rigorous way ?
      – S.H.W
      Nov 20 at 0:52










      up vote
      1
      down vote













      $$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$



      $$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$



      $$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$



      $$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$






      share|cite|improve this answer

























        up vote
        1
        down vote













        $$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$



        $$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$



        $$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$



        $$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          $$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$



          $$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$



          $$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$



          $$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$






          share|cite|improve this answer












          $$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$



          $$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$



          $$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$



          $$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 1:54









          Offlaw

          2649




          2649






















              up vote
              1
              down vote













              Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.






                  share|cite|improve this answer












                  Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 20 at 2:09









                  gb2017

                  944




                  944






























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