Evaluate $lim_{x to 0^+ } x^{x^{x}} - x^x$












4












$begingroup$



Evaluate $$lim_{x to 0^+ } x^{x^{x}} - x^x$$




This is a solved example in my text book but i do not think that the solution is quite correct.



They have essentially used the fact $lim_{xto0^+}x^x$ is 1 and used that to write the term to be evaluated as $$0^1 - 1$$ which gives an answer of -1 The graph indeed gives the limit at $0^+$ as -1.



BUT



We could have used $lim_{xto0^+}x^x$ to evaluate $lim_{x to 0^+ } x^{x^{x}} - x^x$ as $$1^0 - 1$$ which does not give the correct answer.



Is the book's method correct?






limits







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
    $endgroup$
    – MisterRiemann
    Dec 12 '18 at 18:32












  • $begingroup$
    Related.
    $endgroup$
    – Shaun
    Dec 12 '18 at 18:39
















4












$begingroup$



Evaluate $$lim_{x to 0^+ } x^{x^{x}} - x^x$$




This is a solved example in my text book but i do not think that the solution is quite correct.



They have essentially used the fact $lim_{xto0^+}x^x$ is 1 and used that to write the term to be evaluated as $$0^1 - 1$$ which gives an answer of -1 The graph indeed gives the limit at $0^+$ as -1.



BUT



We could have used $lim_{xto0^+}x^x$ to evaluate $lim_{x to 0^+ } x^{x^{x}} - x^x$ as $$1^0 - 1$$ which does not give the correct answer.



Is the book's method correct?






limits







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
    $endgroup$
    – MisterRiemann
    Dec 12 '18 at 18:32












  • $begingroup$
    Related.
    $endgroup$
    – Shaun
    Dec 12 '18 at 18:39














4












4








4





$begingroup$



Evaluate $$lim_{x to 0^+ } x^{x^{x}} - x^x$$




This is a solved example in my text book but i do not think that the solution is quite correct.



They have essentially used the fact $lim_{xto0^+}x^x$ is 1 and used that to write the term to be evaluated as $$0^1 - 1$$ which gives an answer of -1 The graph indeed gives the limit at $0^+$ as -1.



BUT



We could have used $lim_{xto0^+}x^x$ to evaluate $lim_{x to 0^+ } x^{x^{x}} - x^x$ as $$1^0 - 1$$ which does not give the correct answer.



Is the book's method correct?






limits







share|cite|improve this question









$endgroup$





Evaluate $$lim_{x to 0^+ } x^{x^{x}} - x^x$$




This is a solved example in my text book but i do not think that the solution is quite correct.



They have essentially used the fact $lim_{xto0^+}x^x$ is 1 and used that to write the term to be evaluated as $$0^1 - 1$$ which gives an answer of -1 The graph indeed gives the limit at $0^+$ as -1.



BUT



We could have used $lim_{xto0^+}x^x$ to evaluate $lim_{x to 0^+ } x^{x^{x}} - x^x$ as $$1^0 - 1$$ which does not give the correct answer.



Is the book's method correct?






limits




limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 18:30









Avnish KabajAvnish Kabaj

174111




174111








  • 5




    $begingroup$
    Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
    $endgroup$
    – MisterRiemann
    Dec 12 '18 at 18:32












  • $begingroup$
    Related.
    $endgroup$
    – Shaun
    Dec 12 '18 at 18:39














  • 5




    $begingroup$
    Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
    $endgroup$
    – MisterRiemann
    Dec 12 '18 at 18:32












  • $begingroup$
    Related.
    $endgroup$
    – Shaun
    Dec 12 '18 at 18:39








5




5




$begingroup$
Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
$endgroup$
– MisterRiemann
Dec 12 '18 at 18:32






$begingroup$
Note that $$ (x^x)^x not= x^{(x^x)}. $$ (You're looking at the latter, not the former.)
$endgroup$
– MisterRiemann
Dec 12 '18 at 18:32














$begingroup$
Related.
$endgroup$
– Shaun
Dec 12 '18 at 18:39




$begingroup$
Related.
$endgroup$
– Shaun
Dec 12 '18 at 18:39










2 Answers
2






active

oldest

votes


















4












$begingroup$

Note that you solve for the exponent first.



$$a^{b^c} color{red}{neq (a^b)^c = a^{bc}}$$



For example,



$$2^{3^2} = 2^9 = 512 color{red}{neq (2^3)^2 = 2^6 = 64}$$



Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The given solution uses that $x^x to 1$ and



    $$(x)^{(x^{x})} - x^x to 0^1-1=-1$$



    while you are considering



    $$(x^x)^{x} - x^x =x^{(x^2)}-x^xto 1-1=0$$



    indeed





    • $(x)^{(x^{x})}=e^{x^xlog x} to 0$ (since $x^x log x to -infty$)


    but




    • $x^{(x^2)}=e^{x^2log x} to e^0=1$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Note that you solve for the exponent first.



      $$a^{b^c} color{red}{neq (a^b)^c = a^{bc}}$$



      For example,



      $$2^{3^2} = 2^9 = 512 color{red}{neq (2^3)^2 = 2^6 = 64}$$



      Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Note that you solve for the exponent first.



        $$a^{b^c} color{red}{neq (a^b)^c = a^{bc}}$$



        For example,



        $$2^{3^2} = 2^9 = 512 color{red}{neq (2^3)^2 = 2^6 = 64}$$



        Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Note that you solve for the exponent first.



          $$a^{b^c} color{red}{neq (a^b)^c = a^{bc}}$$



          For example,



          $$2^{3^2} = 2^9 = 512 color{red}{neq (2^3)^2 = 2^6 = 64}$$



          Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.






          share|cite|improve this answer









          $endgroup$



          Note that you solve for the exponent first.



          $$a^{b^c} color{red}{neq (a^b)^c = a^{bc}}$$



          For example,



          $$2^{3^2} = 2^9 = 512 color{red}{neq (2^3)^2 = 2^6 = 64}$$



          Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 18:34









          KM101KM101

          6,0351525




          6,0351525























              2












              $begingroup$

              The given solution uses that $x^x to 1$ and



              $$(x)^{(x^{x})} - x^x to 0^1-1=-1$$



              while you are considering



              $$(x^x)^{x} - x^x =x^{(x^2)}-x^xto 1-1=0$$



              indeed





              • $(x)^{(x^{x})}=e^{x^xlog x} to 0$ (since $x^x log x to -infty$)


              but




              • $x^{(x^2)}=e^{x^2log x} to e^0=1$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                The given solution uses that $x^x to 1$ and



                $$(x)^{(x^{x})} - x^x to 0^1-1=-1$$



                while you are considering



                $$(x^x)^{x} - x^x =x^{(x^2)}-x^xto 1-1=0$$



                indeed





                • $(x)^{(x^{x})}=e^{x^xlog x} to 0$ (since $x^x log x to -infty$)


                but




                • $x^{(x^2)}=e^{x^2log x} to e^0=1$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The given solution uses that $x^x to 1$ and



                  $$(x)^{(x^{x})} - x^x to 0^1-1=-1$$



                  while you are considering



                  $$(x^x)^{x} - x^x =x^{(x^2)}-x^xto 1-1=0$$



                  indeed





                  • $(x)^{(x^{x})}=e^{x^xlog x} to 0$ (since $x^x log x to -infty$)


                  but




                  • $x^{(x^2)}=e^{x^2log x} to e^0=1$






                  share|cite|improve this answer











                  $endgroup$



                  The given solution uses that $x^x to 1$ and



                  $$(x)^{(x^{x})} - x^x to 0^1-1=-1$$



                  while you are considering



                  $$(x^x)^{x} - x^x =x^{(x^2)}-x^xto 1-1=0$$



                  indeed





                  • $(x)^{(x^{x})}=e^{x^xlog x} to 0$ (since $x^x log x to -infty$)


                  but




                  • $x^{(x^2)}=e^{x^2log x} to e^0=1$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 18:42

























                  answered Dec 12 '18 at 18:36









                  gimusigimusi

                  92.9k84494




                  92.9k84494






























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