Does the value of the $lim_{x to 0-} x^x = 1$?












2












$begingroup$


I have the following attempt.



Let $x=-y$ then ${y to 0+}$ as ${x to 0-}$.



So, $displaystylelim_{x to 0-} {x}^{x}$= $displaystylelim_{y to 0+} {(-y)}^{(-y)} = displaystylelim_{y to 0+} dfrac{1}{{(-y)}^{y}}= displaystylelim_{y to 0+} dfrac{1}{{(-1)}^{y}.{y}^{y}}=displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}$



Now as, $displaystylelim_{y to 0+} y^y =displaystylelim_{y to 0+} {e}^{yln{y}}
= {e}^{displaystylelim_{y to 0+} yln{y}}={e}^{displaystylelim_{y to 0+} frac{ln{y}}{frac{1}{y}}} = {e}^{displaystylelim_{y to 0+} frac{frac{1}{y}}{{-frac{1}{y^2}}}}
= {e}^{displaystylelim_{y to 0+} {-y}}=e^{0}=1$



Hence $displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}=dfrac{1}{1}=1$



So, $displaystylelim_{x to 0-} {x}^{x}=1$



Is it correct?






complex-analysis limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 24 '18 at 15:32






  • 1




    $begingroup$
    How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 24 '18 at 15:32






  • 2




    $begingroup$
    It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 15:33








  • 1




    $begingroup$
    $x^y$ is not defined if$x<0$ and$y$ is not an integer.
    $endgroup$
    – Bernard
    Dec 24 '18 at 15:40










  • $begingroup$
    Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
    $endgroup$
    – The Pointer
    Dec 24 '18 at 15:57


















2












$begingroup$


I have the following attempt.



Let $x=-y$ then ${y to 0+}$ as ${x to 0-}$.



So, $displaystylelim_{x to 0-} {x}^{x}$= $displaystylelim_{y to 0+} {(-y)}^{(-y)} = displaystylelim_{y to 0+} dfrac{1}{{(-y)}^{y}}= displaystylelim_{y to 0+} dfrac{1}{{(-1)}^{y}.{y}^{y}}=displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}$



Now as, $displaystylelim_{y to 0+} y^y =displaystylelim_{y to 0+} {e}^{yln{y}}
= {e}^{displaystylelim_{y to 0+} yln{y}}={e}^{displaystylelim_{y to 0+} frac{ln{y}}{frac{1}{y}}} = {e}^{displaystylelim_{y to 0+} frac{frac{1}{y}}{{-frac{1}{y^2}}}}
= {e}^{displaystylelim_{y to 0+} {-y}}=e^{0}=1$



Hence $displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}=dfrac{1}{1}=1$



So, $displaystylelim_{x to 0-} {x}^{x}=1$



Is it correct?






complex-analysis limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 24 '18 at 15:32






  • 1




    $begingroup$
    How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 24 '18 at 15:32






  • 2




    $begingroup$
    It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 15:33








  • 1




    $begingroup$
    $x^y$ is not defined if$x<0$ and$y$ is not an integer.
    $endgroup$
    – Bernard
    Dec 24 '18 at 15:40










  • $begingroup$
    Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
    $endgroup$
    – The Pointer
    Dec 24 '18 at 15:57
















2












2








2





$begingroup$


I have the following attempt.



Let $x=-y$ then ${y to 0+}$ as ${x to 0-}$.



So, $displaystylelim_{x to 0-} {x}^{x}$= $displaystylelim_{y to 0+} {(-y)}^{(-y)} = displaystylelim_{y to 0+} dfrac{1}{{(-y)}^{y}}= displaystylelim_{y to 0+} dfrac{1}{{(-1)}^{y}.{y}^{y}}=displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}$



Now as, $displaystylelim_{y to 0+} y^y =displaystylelim_{y to 0+} {e}^{yln{y}}
= {e}^{displaystylelim_{y to 0+} yln{y}}={e}^{displaystylelim_{y to 0+} frac{ln{y}}{frac{1}{y}}} = {e}^{displaystylelim_{y to 0+} frac{frac{1}{y}}{{-frac{1}{y^2}}}}
= {e}^{displaystylelim_{y to 0+} {-y}}=e^{0}=1$



Hence $displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}=dfrac{1}{1}=1$



So, $displaystylelim_{x to 0-} {x}^{x}=1$



Is it correct?






complex-analysis limits







share|cite|improve this question











$endgroup$




I have the following attempt.



Let $x=-y$ then ${y to 0+}$ as ${x to 0-}$.



So, $displaystylelim_{x to 0-} {x}^{x}$= $displaystylelim_{y to 0+} {(-y)}^{(-y)} = displaystylelim_{y to 0+} dfrac{1}{{(-y)}^{y}}= displaystylelim_{y to 0+} dfrac{1}{{(-1)}^{y}.{y}^{y}}=displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}$



Now as, $displaystylelim_{y to 0+} y^y =displaystylelim_{y to 0+} {e}^{yln{y}}
= {e}^{displaystylelim_{y to 0+} yln{y}}={e}^{displaystylelim_{y to 0+} frac{ln{y}}{frac{1}{y}}} = {e}^{displaystylelim_{y to 0+} frac{frac{1}{y}}{{-frac{1}{y^2}}}}
= {e}^{displaystylelim_{y to 0+} {-y}}=e^{0}=1$



Hence $displaystylelim_{y to 0+} dfrac{1}{{y}^{y}}=dfrac{1}{1}=1$



So, $displaystylelim_{x to 0-} {x}^{x}=1$



Is it correct?






complex-analysis limits




complex-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 15:45







Kousik Sett

















asked Dec 24 '18 at 15:30









Kousik SettKousik Sett

1026




1026












  • $begingroup$
    For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 24 '18 at 15:32






  • 1




    $begingroup$
    How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 24 '18 at 15:32






  • 2




    $begingroup$
    It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 15:33








  • 1




    $begingroup$
    $x^y$ is not defined if$x<0$ and$y$ is not an integer.
    $endgroup$
    – Bernard
    Dec 24 '18 at 15:40










  • $begingroup$
    Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
    $endgroup$
    – The Pointer
    Dec 24 '18 at 15:57




















  • $begingroup$
    For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 24 '18 at 15:32






  • 1




    $begingroup$
    How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 24 '18 at 15:32






  • 2




    $begingroup$
    It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 15:33








  • 1




    $begingroup$
    $x^y$ is not defined if$x<0$ and$y$ is not an integer.
    $endgroup$
    – Bernard
    Dec 24 '18 at 15:40










  • $begingroup$
    Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
    $endgroup$
    – The Pointer
    Dec 24 '18 at 15:57


















$begingroup$
For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
$endgroup$
– Dr. Sonnhard Graubner
Dec 24 '18 at 15:32




$begingroup$
For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number.
$endgroup$
– Dr. Sonnhard Graubner
Dec 24 '18 at 15:32




1




1




$begingroup$
How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
$endgroup$
– Aniruddha Deshmukh
Dec 24 '18 at 15:32




$begingroup$
How did you write $left( -1 right)^y = 1$? Don't you think if we take a sequence $dfrac{1}{n}$, $left( -1 right)^{frac{1}{n}}$ may not be defined in $mathbb{R}$?
$endgroup$
– Aniruddha Deshmukh
Dec 24 '18 at 15:32




2




2




$begingroup$
It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:33






$begingroup$
It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(ln(-a)+pi i)}$ your argument should work.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:33






1




1




$begingroup$
$x^y$ is not defined if$x<0$ and$y$ is not an integer.
$endgroup$
– Bernard
Dec 24 '18 at 15:40




$begingroup$
$x^y$ is not defined if$x<0$ and$y$ is not an integer.
$endgroup$
– Bernard
Dec 24 '18 at 15:40












$begingroup$
Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
$endgroup$
– The Pointer
Dec 24 '18 at 15:57






$begingroup$
Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX.
$endgroup$
– The Pointer
Dec 24 '18 at 15:57












2 Answers
2






active

oldest

votes


















2












$begingroup$

For complex values of $z$ and $w$, we have by definition



$$begin{align}
z^w&=e^{wlog(z)}\\
&=e^{wtext{Log}(|z|)+iwarg(z)}tag1
end{align}$$



where $text{Log}$ is the logarithm function of real variables and $arg(z)$ is the multi-valued argument of $z$.



Using $(1)$ reveals for $xin mathbb{R}$ and $x<0$



$$begin{align}
lim_{xto 0^-}x^x&=lim_{xto 0^-}e^{xtext{Log}(|x|)+ixarg(x)}\\
&=lim_{xto 0^-}x^{|x|}e^{ix(2n+1)pi}\\
&=1
end{align}$$



as was to be shown!






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $x<0$. It holds that $x^x=e^{xlog x} = e^{x (log(-x)+pi i)} = e^{xlog(-x)}(cos(pi x) + i sin (pi x))$. And I think you can complete the details.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
      $endgroup$
      – Mark Viola
      Dec 24 '18 at 16:17












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For complex values of $z$ and $w$, we have by definition



    $$begin{align}
    z^w&=e^{wlog(z)}\\
    &=e^{wtext{Log}(|z|)+iwarg(z)}tag1
    end{align}$$



    where $text{Log}$ is the logarithm function of real variables and $arg(z)$ is the multi-valued argument of $z$.



    Using $(1)$ reveals for $xin mathbb{R}$ and $x<0$



    $$begin{align}
    lim_{xto 0^-}x^x&=lim_{xto 0^-}e^{xtext{Log}(|x|)+ixarg(x)}\\
    &=lim_{xto 0^-}x^{|x|}e^{ix(2n+1)pi}\\
    &=1
    end{align}$$



    as was to be shown!






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      For complex values of $z$ and $w$, we have by definition



      $$begin{align}
      z^w&=e^{wlog(z)}\\
      &=e^{wtext{Log}(|z|)+iwarg(z)}tag1
      end{align}$$



      where $text{Log}$ is the logarithm function of real variables and $arg(z)$ is the multi-valued argument of $z$.



      Using $(1)$ reveals for $xin mathbb{R}$ and $x<0$



      $$begin{align}
      lim_{xto 0^-}x^x&=lim_{xto 0^-}e^{xtext{Log}(|x|)+ixarg(x)}\\
      &=lim_{xto 0^-}x^{|x|}e^{ix(2n+1)pi}\\
      &=1
      end{align}$$



      as was to be shown!






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        For complex values of $z$ and $w$, we have by definition



        $$begin{align}
        z^w&=e^{wlog(z)}\\
        &=e^{wtext{Log}(|z|)+iwarg(z)}tag1
        end{align}$$



        where $text{Log}$ is the logarithm function of real variables and $arg(z)$ is the multi-valued argument of $z$.



        Using $(1)$ reveals for $xin mathbb{R}$ and $x<0$



        $$begin{align}
        lim_{xto 0^-}x^x&=lim_{xto 0^-}e^{xtext{Log}(|x|)+ixarg(x)}\\
        &=lim_{xto 0^-}x^{|x|}e^{ix(2n+1)pi}\\
        &=1
        end{align}$$



        as was to be shown!






        share|cite|improve this answer









        $endgroup$



        For complex values of $z$ and $w$, we have by definition



        $$begin{align}
        z^w&=e^{wlog(z)}\\
        &=e^{wtext{Log}(|z|)+iwarg(z)}tag1
        end{align}$$



        where $text{Log}$ is the logarithm function of real variables and $arg(z)$ is the multi-valued argument of $z$.



        Using $(1)$ reveals for $xin mathbb{R}$ and $x<0$



        $$begin{align}
        lim_{xto 0^-}x^x&=lim_{xto 0^-}e^{xtext{Log}(|x|)+ixarg(x)}\\
        &=lim_{xto 0^-}x^{|x|}e^{ix(2n+1)pi}\\
        &=1
        end{align}$$



        as was to be shown!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 16:14









        Mark ViolaMark Viola

        134k1278177




        134k1278177























            0












            $begingroup$

            Let $x<0$. It holds that $x^x=e^{xlog x} = e^{x (log(-x)+pi i)} = e^{xlog(-x)}(cos(pi x) + i sin (pi x))$. And I think you can complete the details.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
              $endgroup$
              – Mark Viola
              Dec 24 '18 at 16:17
















            0












            $begingroup$

            Let $x<0$. It holds that $x^x=e^{xlog x} = e^{x (log(-x)+pi i)} = e^{xlog(-x)}(cos(pi x) + i sin (pi x))$. And I think you can complete the details.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
              $endgroup$
              – Mark Viola
              Dec 24 '18 at 16:17














            0












            0








            0





            $begingroup$

            Let $x<0$. It holds that $x^x=e^{xlog x} = e^{x (log(-x)+pi i)} = e^{xlog(-x)}(cos(pi x) + i sin (pi x))$. And I think you can complete the details.






            share|cite|improve this answer









            $endgroup$



            Let $x<0$. It holds that $x^x=e^{xlog x} = e^{x (log(-x)+pi i)} = e^{xlog(-x)}(cos(pi x) + i sin (pi x))$. And I think you can complete the details.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 16:13









            MirceaMircea

            1736




            1736












            • $begingroup$
              Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
              $endgroup$
              – Mark Viola
              Dec 24 '18 at 16:17


















            • $begingroup$
              Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
              $endgroup$
              – Mark Viola
              Dec 24 '18 at 16:17
















            $begingroup$
            Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
            $endgroup$
            – Mark Viola
            Dec 24 '18 at 16:17




            $begingroup$
            Note that for $x<0$, $log(x)=text{Log}(|x|)+i(2n+1)pi$, for $nin mathbb{Z}$.
            $endgroup$
            – Mark Viola
            Dec 24 '18 at 16:17


















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