How do you simplify $lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}$?












0












$begingroup$


I'm having an issue trying to figure out parameter to check for what parameter $b$ function is continuous.



$$
begin{cases}
arctanleft(frac{1}{x-1}right) &text{for $x<1$} \
a &text{for $x=1$} \
bcdot e^{ln(x)cdot ln(ln(x))} &text{for $x > 1$}
end{cases}
$$



In order to solve that, I need to figure out answer for
$$
frac{-pi/2}b = lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}
$$



I wanted to use L'Hôpital's rule, but I'm having trouble substituting indeterminate form $e^{0cdot (-infty)}$ to $frac 00$.






calculus limits







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  • 1




    $begingroup$
    Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
    $endgroup$
    – Christoph
    Dec 18 '18 at 16:32












  • $begingroup$
    Mathematica yields $1$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 16:35
















0












$begingroup$


I'm having an issue trying to figure out parameter to check for what parameter $b$ function is continuous.



$$
begin{cases}
arctanleft(frac{1}{x-1}right) &text{for $x<1$} \
a &text{for $x=1$} \
bcdot e^{ln(x)cdot ln(ln(x))} &text{for $x > 1$}
end{cases}
$$



In order to solve that, I need to figure out answer for
$$
frac{-pi/2}b = lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}
$$



I wanted to use L'Hôpital's rule, but I'm having trouble substituting indeterminate form $e^{0cdot (-infty)}$ to $frac 00$.






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
    $endgroup$
    – Christoph
    Dec 18 '18 at 16:32












  • $begingroup$
    Mathematica yields $1$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 16:35














0












0








0





$begingroup$


I'm having an issue trying to figure out parameter to check for what parameter $b$ function is continuous.



$$
begin{cases}
arctanleft(frac{1}{x-1}right) &text{for $x<1$} \
a &text{for $x=1$} \
bcdot e^{ln(x)cdot ln(ln(x))} &text{for $x > 1$}
end{cases}
$$



In order to solve that, I need to figure out answer for
$$
frac{-pi/2}b = lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}
$$



I wanted to use L'Hôpital's rule, but I'm having trouble substituting indeterminate form $e^{0cdot (-infty)}$ to $frac 00$.






calculus limits







share|cite|improve this question











$endgroup$




I'm having an issue trying to figure out parameter to check for what parameter $b$ function is continuous.



$$
begin{cases}
arctanleft(frac{1}{x-1}right) &text{for $x<1$} \
a &text{for $x=1$} \
bcdot e^{ln(x)cdot ln(ln(x))} &text{for $x > 1$}
end{cases}
$$



In order to solve that, I need to figure out answer for
$$
frac{-pi/2}b = lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}
$$



I wanted to use L'Hôpital's rule, but I'm having trouble substituting indeterminate form $e^{0cdot (-infty)}$ to $frac 00$.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 16:31









Christoph

12.5k1642




12.5k1642










asked Dec 18 '18 at 16:26









A. OsuchA. Osuch

1




1








  • 1




    $begingroup$
    Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
    $endgroup$
    – Christoph
    Dec 18 '18 at 16:32












  • $begingroup$
    Mathematica yields $1$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 16:35














  • 1




    $begingroup$
    Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
    $endgroup$
    – Christoph
    Dec 18 '18 at 16:32












  • $begingroup$
    Mathematica yields $1$.
    $endgroup$
    – David G. Stork
    Dec 18 '18 at 16:35








1




1




$begingroup$
Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
$endgroup$
– Christoph
Dec 18 '18 at 16:32






$begingroup$
Hi and welcome to MSE. I edited your post to improve the formatting using MathJax. Please have a look at the tutorial so you can format your future posts nicely on your own.
$endgroup$
– Christoph
Dec 18 '18 at 16:32














$begingroup$
Mathematica yields $1$.
$endgroup$
– David G. Stork
Dec 18 '18 at 16:35




$begingroup$
Mathematica yields $1$.
$endgroup$
– David G. Stork
Dec 18 '18 at 16:35










5 Answers
5






active

oldest

votes


















3












$begingroup$

Since the exponential function $xmapsto e^x$ is continuous, this boils down to
$$
lim_{xto 1+} ln(x)cdot ln(ln(x)).
$$
In order to apply L'Hôpital's rule, you can write this as
$$
frac{ln(x)}{ frac{1}{ln(ln(x))} }
qquadtext{or}qquad
frac{ln(ln(x))}{ frac{1}{ln(x)} }.
$$




You can also "simplify" this further by noting that
$$
ln(x)cdotln(ln(x)) = lnleft(ln(x)^{ln(x)}right).
$$
Again, since $ln$ is continuous you only need to calculate the limit of $ln(x)^{ln(x)}$. However, calculating a limit of a function of the form $f(x)^{g(x)}$ you would usually take the logarithm, so you get back to where you started.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    If the limit exists, $$lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}=e^{lim_{xto 1+}ln(x)cdot ln(ln(x))}$$
    As you noted, this is a $0cdotinfty$ type. What you can do it is write it as $$ln(x)cdot ln(ln(x))=frac{ln(x)}{frac1{ln(ln(x))}}$$
    This is $0/0$ type of limit. Or you can also get it into $infty/infty$ type $$frac{ln(ln(x))}{frac1{ln(x)}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
      $endgroup$
      – A. Osuch
      Dec 18 '18 at 16:45












    • $begingroup$
      Note my first equation. You need to calculate the $0/0$ first
      $endgroup$
      – Andrei
      Dec 18 '18 at 16:47










    • $begingroup$
      Right, I missed it. Thanks a lot.
      $endgroup$
      – A. Osuch
      Dec 18 '18 at 16:52



















    0












    $begingroup$

    Mathematica gives (symbolically) $1$, as supported by the graph:



    enter image description here






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Let $t=ln x$ then $displaystyle lim_{xto 1^+}ln x lnln x=lim_{tto 0^+}t ln t=0^-$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Hint: You have



        $$lim_{x to 1^+} e^{ln(x)cdot {ln((ln x))}} = e^{lim_limits{x to 1^+}ln(x)cdot {ln((ln x))}}$$



        So you need to find



        $$lim_{x to 1^+}ln (x)cdot ln(ln (x))$$



        which is in the form $0(-infty)$. You can rewrite $ln(x)cdotln(ln x)$ to apply L’Hôpital’s Rule, such as by



        $$frac{ln x}{frac{1}{ln(ln(x))}}$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I think it should be $(ln x)^{ln x}$
          $endgroup$
          – Shubham Johri
          Dec 18 '18 at 16:45










        • $begingroup$
          Yes, corrected.
          $endgroup$
          – KM101
          Dec 18 '18 at 16:47










        • $begingroup$
          Did you forget to get rid of the $ln$ when cancelling it with the exponential?
          $endgroup$
          – Christoph
          Dec 18 '18 at 18:45










        • $begingroup$
          Oops, I’ve corrected it.
          $endgroup$
          – KM101
          Dec 18 '18 at 18:54











        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Since the exponential function $xmapsto e^x$ is continuous, this boils down to
        $$
        lim_{xto 1+} ln(x)cdot ln(ln(x)).
        $$
        In order to apply L'Hôpital's rule, you can write this as
        $$
        frac{ln(x)}{ frac{1}{ln(ln(x))} }
        qquadtext{or}qquad
        frac{ln(ln(x))}{ frac{1}{ln(x)} }.
        $$




        You can also "simplify" this further by noting that
        $$
        ln(x)cdotln(ln(x)) = lnleft(ln(x)^{ln(x)}right).
        $$
        Again, since $ln$ is continuous you only need to calculate the limit of $ln(x)^{ln(x)}$. However, calculating a limit of a function of the form $f(x)^{g(x)}$ you would usually take the logarithm, so you get back to where you started.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Since the exponential function $xmapsto e^x$ is continuous, this boils down to
          $$
          lim_{xto 1+} ln(x)cdot ln(ln(x)).
          $$
          In order to apply L'Hôpital's rule, you can write this as
          $$
          frac{ln(x)}{ frac{1}{ln(ln(x))} }
          qquadtext{or}qquad
          frac{ln(ln(x))}{ frac{1}{ln(x)} }.
          $$




          You can also "simplify" this further by noting that
          $$
          ln(x)cdotln(ln(x)) = lnleft(ln(x)^{ln(x)}right).
          $$
          Again, since $ln$ is continuous you only need to calculate the limit of $ln(x)^{ln(x)}$. However, calculating a limit of a function of the form $f(x)^{g(x)}$ you would usually take the logarithm, so you get back to where you started.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Since the exponential function $xmapsto e^x$ is continuous, this boils down to
            $$
            lim_{xto 1+} ln(x)cdot ln(ln(x)).
            $$
            In order to apply L'Hôpital's rule, you can write this as
            $$
            frac{ln(x)}{ frac{1}{ln(ln(x))} }
            qquadtext{or}qquad
            frac{ln(ln(x))}{ frac{1}{ln(x)} }.
            $$




            You can also "simplify" this further by noting that
            $$
            ln(x)cdotln(ln(x)) = lnleft(ln(x)^{ln(x)}right).
            $$
            Again, since $ln$ is continuous you only need to calculate the limit of $ln(x)^{ln(x)}$. However, calculating a limit of a function of the form $f(x)^{g(x)}$ you would usually take the logarithm, so you get back to where you started.






            share|cite|improve this answer











            $endgroup$



            Since the exponential function $xmapsto e^x$ is continuous, this boils down to
            $$
            lim_{xto 1+} ln(x)cdot ln(ln(x)).
            $$
            In order to apply L'Hôpital's rule, you can write this as
            $$
            frac{ln(x)}{ frac{1}{ln(ln(x))} }
            qquadtext{or}qquad
            frac{ln(ln(x))}{ frac{1}{ln(x)} }.
            $$




            You can also "simplify" this further by noting that
            $$
            ln(x)cdotln(ln(x)) = lnleft(ln(x)^{ln(x)}right).
            $$
            Again, since $ln$ is continuous you only need to calculate the limit of $ln(x)^{ln(x)}$. However, calculating a limit of a function of the form $f(x)^{g(x)}$ you would usually take the logarithm, so you get back to where you started.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 18:48

























            answered Dec 18 '18 at 16:38









            ChristophChristoph

            12.5k1642




            12.5k1642























                2












                $begingroup$

                If the limit exists, $$lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}=e^{lim_{xto 1+}ln(x)cdot ln(ln(x))}$$
                As you noted, this is a $0cdotinfty$ type. What you can do it is write it as $$ln(x)cdot ln(ln(x))=frac{ln(x)}{frac1{ln(ln(x))}}$$
                This is $0/0$ type of limit. Or you can also get it into $infty/infty$ type $$frac{ln(ln(x))}{frac1{ln(x)}}$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:45












                • $begingroup$
                  Note my first equation. You need to calculate the $0/0$ first
                  $endgroup$
                  – Andrei
                  Dec 18 '18 at 16:47










                • $begingroup$
                  Right, I missed it. Thanks a lot.
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:52
















                2












                $begingroup$

                If the limit exists, $$lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}=e^{lim_{xto 1+}ln(x)cdot ln(ln(x))}$$
                As you noted, this is a $0cdotinfty$ type. What you can do it is write it as $$ln(x)cdot ln(ln(x))=frac{ln(x)}{frac1{ln(ln(x))}}$$
                This is $0/0$ type of limit. Or you can also get it into $infty/infty$ type $$frac{ln(ln(x))}{frac1{ln(x)}}$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:45












                • $begingroup$
                  Note my first equation. You need to calculate the $0/0$ first
                  $endgroup$
                  – Andrei
                  Dec 18 '18 at 16:47










                • $begingroup$
                  Right, I missed it. Thanks a lot.
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:52














                2












                2








                2





                $begingroup$

                If the limit exists, $$lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}=e^{lim_{xto 1+}ln(x)cdot ln(ln(x))}$$
                As you noted, this is a $0cdotinfty$ type. What you can do it is write it as $$ln(x)cdot ln(ln(x))=frac{ln(x)}{frac1{ln(ln(x))}}$$
                This is $0/0$ type of limit. Or you can also get it into $infty/infty$ type $$frac{ln(ln(x))}{frac1{ln(x)}}$$






                share|cite|improve this answer









                $endgroup$



                If the limit exists, $$lim_{xto 1+} e^{ln(x)cdot ln(ln(x))}=e^{lim_{xto 1+}ln(x)cdot ln(ln(x))}$$
                As you noted, this is a $0cdotinfty$ type. What you can do it is write it as $$ln(x)cdot ln(ln(x))=frac{ln(x)}{frac1{ln(ln(x))}}$$
                This is $0/0$ type of limit. Or you can also get it into $infty/infty$ type $$frac{ln(ln(x))}{frac1{ln(x)}}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 16:38









                AndreiAndrei

                13.1k21230




                13.1k21230












                • $begingroup$
                  I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:45












                • $begingroup$
                  Note my first equation. You need to calculate the $0/0$ first
                  $endgroup$
                  – Andrei
                  Dec 18 '18 at 16:47










                • $begingroup$
                  Right, I missed it. Thanks a lot.
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:52


















                • $begingroup$
                  I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:45












                • $begingroup$
                  Note my first equation. You need to calculate the $0/0$ first
                  $endgroup$
                  – Andrei
                  Dec 18 '18 at 16:47










                • $begingroup$
                  Right, I missed it. Thanks a lot.
                  $endgroup$
                  – A. Osuch
                  Dec 18 '18 at 16:52
















                $begingroup$
                I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                $endgroup$
                – A. Osuch
                Dec 18 '18 at 16:45






                $begingroup$
                I'm probably misunderstanding something, but it seems to me that it would lead me to e^(0/0) which is not the same as 0/0, so i can't use L'Hôpital's rule. Am I wrong?
                $endgroup$
                – A. Osuch
                Dec 18 '18 at 16:45














                $begingroup$
                Note my first equation. You need to calculate the $0/0$ first
                $endgroup$
                – Andrei
                Dec 18 '18 at 16:47




                $begingroup$
                Note my first equation. You need to calculate the $0/0$ first
                $endgroup$
                – Andrei
                Dec 18 '18 at 16:47












                $begingroup$
                Right, I missed it. Thanks a lot.
                $endgroup$
                – A. Osuch
                Dec 18 '18 at 16:52




                $begingroup$
                Right, I missed it. Thanks a lot.
                $endgroup$
                – A. Osuch
                Dec 18 '18 at 16:52











                0












                $begingroup$

                Mathematica gives (symbolically) $1$, as supported by the graph:



                enter image description here






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Mathematica gives (symbolically) $1$, as supported by the graph:



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Mathematica gives (symbolically) $1$, as supported by the graph:



                    enter image description here






                    share|cite|improve this answer









                    $endgroup$



                    Mathematica gives (symbolically) $1$, as supported by the graph:



                    enter image description here







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 18 '18 at 16:37









                    David G. StorkDavid G. Stork

                    11.1k41432




                    11.1k41432























                        0












                        $begingroup$

                        Let $t=ln x$ then $displaystyle lim_{xto 1^+}ln x lnln x=lim_{tto 0^+}t ln t=0^-$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Let $t=ln x$ then $displaystyle lim_{xto 1^+}ln x lnln x=lim_{tto 0^+}t ln t=0^-$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Let $t=ln x$ then $displaystyle lim_{xto 1^+}ln x lnln x=lim_{tto 0^+}t ln t=0^-$.






                            share|cite|improve this answer









                            $endgroup$



                            Let $t=ln x$ then $displaystyle lim_{xto 1^+}ln x lnln x=lim_{tto 0^+}t ln t=0^-$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 18 '18 at 16:39









                            BPPBPP

                            2,168927




                            2,168927























                                0












                                $begingroup$

                                Hint: You have



                                $$lim_{x to 1^+} e^{ln(x)cdot {ln((ln x))}} = e^{lim_limits{x to 1^+}ln(x)cdot {ln((ln x))}}$$



                                So you need to find



                                $$lim_{x to 1^+}ln (x)cdot ln(ln (x))$$



                                which is in the form $0(-infty)$. You can rewrite $ln(x)cdotln(ln x)$ to apply L’Hôpital’s Rule, such as by



                                $$frac{ln x}{frac{1}{ln(ln(x))}}$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  I think it should be $(ln x)^{ln x}$
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 18 '18 at 16:45










                                • $begingroup$
                                  Yes, corrected.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 16:47










                                • $begingroup$
                                  Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                  $endgroup$
                                  – Christoph
                                  Dec 18 '18 at 18:45










                                • $begingroup$
                                  Oops, I’ve corrected it.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 18:54
















                                0












                                $begingroup$

                                Hint: You have



                                $$lim_{x to 1^+} e^{ln(x)cdot {ln((ln x))}} = e^{lim_limits{x to 1^+}ln(x)cdot {ln((ln x))}}$$



                                So you need to find



                                $$lim_{x to 1^+}ln (x)cdot ln(ln (x))$$



                                which is in the form $0(-infty)$. You can rewrite $ln(x)cdotln(ln x)$ to apply L’Hôpital’s Rule, such as by



                                $$frac{ln x}{frac{1}{ln(ln(x))}}$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  I think it should be $(ln x)^{ln x}$
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 18 '18 at 16:45










                                • $begingroup$
                                  Yes, corrected.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 16:47










                                • $begingroup$
                                  Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                  $endgroup$
                                  – Christoph
                                  Dec 18 '18 at 18:45










                                • $begingroup$
                                  Oops, I’ve corrected it.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 18:54














                                0












                                0








                                0





                                $begingroup$

                                Hint: You have



                                $$lim_{x to 1^+} e^{ln(x)cdot {ln((ln x))}} = e^{lim_limits{x to 1^+}ln(x)cdot {ln((ln x))}}$$



                                So you need to find



                                $$lim_{x to 1^+}ln (x)cdot ln(ln (x))$$



                                which is in the form $0(-infty)$. You can rewrite $ln(x)cdotln(ln x)$ to apply L’Hôpital’s Rule, such as by



                                $$frac{ln x}{frac{1}{ln(ln(x))}}$$






                                share|cite|improve this answer











                                $endgroup$



                                Hint: You have



                                $$lim_{x to 1^+} e^{ln(x)cdot {ln((ln x))}} = e^{lim_limits{x to 1^+}ln(x)cdot {ln((ln x))}}$$



                                So you need to find



                                $$lim_{x to 1^+}ln (x)cdot ln(ln (x))$$



                                which is in the form $0(-infty)$. You can rewrite $ln(x)cdotln(ln x)$ to apply L’Hôpital’s Rule, such as by



                                $$frac{ln x}{frac{1}{ln(ln(x))}}$$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 18 '18 at 18:50

























                                answered Dec 18 '18 at 16:42









                                KM101KM101

                                6,0901525




                                6,0901525












                                • $begingroup$
                                  I think it should be $(ln x)^{ln x}$
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 18 '18 at 16:45










                                • $begingroup$
                                  Yes, corrected.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 16:47










                                • $begingroup$
                                  Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                  $endgroup$
                                  – Christoph
                                  Dec 18 '18 at 18:45










                                • $begingroup$
                                  Oops, I’ve corrected it.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 18:54


















                                • $begingroup$
                                  I think it should be $(ln x)^{ln x}$
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 18 '18 at 16:45










                                • $begingroup$
                                  Yes, corrected.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 16:47










                                • $begingroup$
                                  Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                  $endgroup$
                                  – Christoph
                                  Dec 18 '18 at 18:45










                                • $begingroup$
                                  Oops, I’ve corrected it.
                                  $endgroup$
                                  – KM101
                                  Dec 18 '18 at 18:54
















                                $begingroup$
                                I think it should be $(ln x)^{ln x}$
                                $endgroup$
                                – Shubham Johri
                                Dec 18 '18 at 16:45




                                $begingroup$
                                I think it should be $(ln x)^{ln x}$
                                $endgroup$
                                – Shubham Johri
                                Dec 18 '18 at 16:45












                                $begingroup$
                                Yes, corrected.
                                $endgroup$
                                – KM101
                                Dec 18 '18 at 16:47




                                $begingroup$
                                Yes, corrected.
                                $endgroup$
                                – KM101
                                Dec 18 '18 at 16:47












                                $begingroup$
                                Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                $endgroup$
                                – Christoph
                                Dec 18 '18 at 18:45




                                $begingroup$
                                Did you forget to get rid of the $ln$ when cancelling it with the exponential?
                                $endgroup$
                                – Christoph
                                Dec 18 '18 at 18:45












                                $begingroup$
                                Oops, I’ve corrected it.
                                $endgroup$
                                – KM101
                                Dec 18 '18 at 18:54




                                $begingroup$
                                Oops, I’ve corrected it.
                                $endgroup$
                                – KM101
                                Dec 18 '18 at 18:54


















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