What is the technique where you perform a “u-substitution” on a limit called?












-1












$begingroup$


I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:



$
lim_{x to 0^+}x^{x^2}$



What you can do to solve this is: set $u = x^2$



Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:



$lim_{x to 0^+}u$



Which gives zero since $0^2 = 0$



And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):



$lim_{u to 0^+}x^u$



And then you get $1$ since anything to the power of zero is $1$.



Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.



Any help is appreciated.






calculus limits







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:



    $
    lim_{x to 0^+}x^{x^2}$



    What you can do to solve this is: set $u = x^2$



    Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:



    $lim_{x to 0^+}u$



    Which gives zero since $0^2 = 0$



    And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):



    $lim_{u to 0^+}x^u$



    And then you get $1$ since anything to the power of zero is $1$.



    Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.



    Any help is appreciated.






    calculus limits







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:



      $
      lim_{x to 0^+}x^{x^2}$



      What you can do to solve this is: set $u = x^2$



      Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:



      $lim_{x to 0^+}u$



      Which gives zero since $0^2 = 0$



      And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):



      $lim_{u to 0^+}x^u$



      And then you get $1$ since anything to the power of zero is $1$.



      Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.



      Any help is appreciated.






      calculus limits







      share|cite|improve this question











      $endgroup$




      I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:



      $
      lim_{x to 0^+}x^{x^2}$



      What you can do to solve this is: set $u = x^2$



      Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:



      $lim_{x to 0^+}u$



      Which gives zero since $0^2 = 0$



      And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):



      $lim_{u to 0^+}x^u$



      And then you get $1$ since anything to the power of zero is $1$.



      Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.



      Any help is appreciated.






      calculus limits




      calculus limits






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 21:31









      Andrei

      12.4k21128




      12.4k21128










      asked Dec 11 '18 at 21:28









      James RonaldJames Ronald

      1237




      1237






















          1 Answer
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          0












          $begingroup$

          Why don't you calculate the logarithm of the limit first:



          $$ln lim_{x to 0^+}x^{x^2}=
          lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
            $endgroup$
            – James Ronald
            Dec 11 '18 at 21:45








          • 2




            $begingroup$
            It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
            $endgroup$
            – Andrei
            Dec 11 '18 at 21:51











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          active

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          0












          $begingroup$

          Why don't you calculate the logarithm of the limit first:



          $$ln lim_{x to 0^+}x^{x^2}=
          lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
            $endgroup$
            – James Ronald
            Dec 11 '18 at 21:45








          • 2




            $begingroup$
            It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
            $endgroup$
            – Andrei
            Dec 11 '18 at 21:51
















          0












          $begingroup$

          Why don't you calculate the logarithm of the limit first:



          $$ln lim_{x to 0^+}x^{x^2}=
          lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
            $endgroup$
            – James Ronald
            Dec 11 '18 at 21:45








          • 2




            $begingroup$
            It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
            $endgroup$
            – Andrei
            Dec 11 '18 at 21:51














          0












          0








          0





          $begingroup$

          Why don't you calculate the logarithm of the limit first:



          $$ln lim_{x to 0^+}x^{x^2}=
          lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$






          share|cite|improve this answer









          $endgroup$



          Why don't you calculate the logarithm of the limit first:



          $$ln lim_{x to 0^+}x^{x^2}=
          lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 21:39









          AndreiAndrei

          12.4k21128




          12.4k21128












          • $begingroup$
            Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
            $endgroup$
            – James Ronald
            Dec 11 '18 at 21:45








          • 2




            $begingroup$
            It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
            $endgroup$
            – Andrei
            Dec 11 '18 at 21:51


















          • $begingroup$
            Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
            $endgroup$
            – James Ronald
            Dec 11 '18 at 21:45








          • 2




            $begingroup$
            It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
            $endgroup$
            – Andrei
            Dec 11 '18 at 21:51
















          $begingroup$
          Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
          $endgroup$
          – James Ronald
          Dec 11 '18 at 21:45






          $begingroup$
          Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
          $endgroup$
          – James Ronald
          Dec 11 '18 at 21:45






          2




          2




          $begingroup$
          It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
          $endgroup$
          – Andrei
          Dec 11 '18 at 21:51




          $begingroup$
          It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
          $endgroup$
          – Andrei
          Dec 11 '18 at 21:51


















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