Evaluate $lim_xrightarrow0^+-xlogx$

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I have the following limit:



$$lim_xrightarrow0^+-xlogx$$



Since this leads to an indeterminate form $[0cdotinfty]$, I change the variables:



$$t=frac 1 x$$



$$lim_trightarrow+infty-fraclog(frac1t)t$$



The denominator grows faster than the nominator so the fraction tends to $0^+$. But since there is a minus there, I flip the value which becomes $0^-$. According to my textbook this is wrong. The limit is supposed to be $0^+$. Any hints on why my approach to get the $pm$ is wrong?




limits




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    up vote
    1
    down vote

    favorite












    I have the following limit:



    $$lim_xrightarrow0^+-xlogx$$



    Since this leads to an indeterminate form $[0cdotinfty]$, I change the variables:



    $$t=frac 1 x$$



    $$lim_trightarrow+infty-fraclog(frac1t)t$$



    The denominator grows faster than the nominator so the fraction tends to $0^+$. But since there is a minus there, I flip the value which becomes $0^-$. According to my textbook this is wrong. The limit is supposed to be $0^+$. Any hints on why my approach to get the $pm$ is wrong?




    limits




    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have the following limit:



      $$lim_xrightarrow0^+-xlogx$$



      Since this leads to an indeterminate form $[0cdotinfty]$, I change the variables:



      $$t=frac 1 x$$



      $$lim_trightarrow+infty-fraclog(frac1t)t$$



      The denominator grows faster than the nominator so the fraction tends to $0^+$. But since there is a minus there, I flip the value which becomes $0^-$. According to my textbook this is wrong. The limit is supposed to be $0^+$. Any hints on why my approach to get the $pm$ is wrong?




      limits




      share|cite|improve this question












      I have the following limit:



      $$lim_xrightarrow0^+-xlogx$$



      Since this leads to an indeterminate form $[0cdotinfty]$, I change the variables:



      $$t=frac 1 x$$



      $$lim_trightarrow+infty-fraclog(frac1t)t$$



      The denominator grows faster than the nominator so the fraction tends to $0^+$. But since there is a minus there, I flip the value which becomes $0^-$. According to my textbook this is wrong. The limit is supposed to be $0^+$. Any hints on why my approach to get the $pm$ is wrong?





      limits





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      asked Aug 21 at 8:17









      Cesare

      587210




      587210




















          2 Answers
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          You are wrong since $$-log1/x=log x$$therefore $$lim_ttoinftydfraclog tt=0^+$$






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          • It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
            – Mostafa Ayaz
            Aug 21 at 8:30

















          up vote
          2
          down vote













          Hint:



          $$logleft(frac1tright)=-log t$$






          share|cite|improve this answer




















          • That's right, thanks
            – Cesare
            Aug 21 at 8:25










          • You are welcome off course.
            – drhab
            Aug 21 at 8:25










          Your Answer




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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          You are wrong since $$-log1/x=log x$$therefore $$lim_ttoinftydfraclog tt=0^+$$






          share|cite|improve this answer




















          • It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
            – Mostafa Ayaz
            Aug 21 at 8:30














          up vote
          1
          down vote



          accepted










          You are wrong since $$-log1/x=log x$$therefore $$lim_ttoinftydfraclog tt=0^+$$






          share|cite|improve this answer




















          • It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
            – Mostafa Ayaz
            Aug 21 at 8:30












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You are wrong since $$-log1/x=log x$$therefore $$lim_ttoinftydfraclog tt=0^+$$






          share|cite|improve this answer












          You are wrong since $$-log1/x=log x$$therefore $$lim_ttoinftydfraclog tt=0^+$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 8:23









          Mostafa Ayaz

          9,9833730




          9,9833730











          • It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
            – Mostafa Ayaz
            Aug 21 at 8:30
















          • It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
            – Mostafa Ayaz
            Aug 21 at 8:30















          It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
          – Mostafa Ayaz
          Aug 21 at 8:30




          It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here
          – Mostafa Ayaz
          Aug 21 at 8:30










          up vote
          2
          down vote













          Hint:



          $$logleft(frac1tright)=-log t$$






          share|cite|improve this answer




















          • That's right, thanks
            – Cesare
            Aug 21 at 8:25










          • You are welcome off course.
            – drhab
            Aug 21 at 8:25














          up vote
          2
          down vote













          Hint:



          $$logleft(frac1tright)=-log t$$






          share|cite|improve this answer




















          • That's right, thanks
            – Cesare
            Aug 21 at 8:25










          • You are welcome off course.
            – drhab
            Aug 21 at 8:25












          up vote
          2
          down vote










          up vote
          2
          down vote









          Hint:



          $$logleft(frac1tright)=-log t$$






          share|cite|improve this answer












          Hint:



          $$logleft(frac1tright)=-log t$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 8:22









          drhab

          87.8k541119




          87.8k541119











          • That's right, thanks
            – Cesare
            Aug 21 at 8:25










          • You are welcome off course.
            – drhab
            Aug 21 at 8:25
















          • That's right, thanks
            – Cesare
            Aug 21 at 8:25










          • You are welcome off course.
            – drhab
            Aug 21 at 8:25















          That's right, thanks
          – Cesare
          Aug 21 at 8:25




          That's right, thanks
          – Cesare
          Aug 21 at 8:25












          You are welcome off course.
          – drhab
          Aug 21 at 8:25




          You are welcome off course.
          – drhab
          Aug 21 at 8:25












           

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